Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 8 Remainder And Factor Theorems

Question 1. Find, in each case, the remainder when:
(i) \( x^4 - 3x^2 + 2x + 1 \) is divided by \( x - 1 \)
(ii) \( x^3 + 3x^2 - 12x + 4 \) is divided by \( x - 2 \)
(iii) \( x^4 + 1 \) is divided by \( x + 1 \)
Answer: By remainder theorem we know that when a polynomial f(x) is divided by \( x - a \), then the remainder is f(a).

(i) f(x) = \( x^4 - 3x^2 + 2x + 1 \)
Remainder = f(1) = \( (1)^4 - 3(1)^2 + 2(1) + 1 = 1 - 3 + 2 + 1 = 1 \)

(ii) f(x) = \( x^3 + 3x^2 - 12x + 4 \)
Remainder = f(2) = \( (2)^3 + 3(2)^2 - 12(2) + 4 = 8 + 12 - 24 + 4 = 0 \)

(iii) f(x) = \( x^4 + 1 \)
Remainder = f(-1) = \( (-1)^4 + 1 = 1 + 1 = 2 \)
In simple words: To find remainder, put the value of x into the polynomial. If you divide by \( x - 1 \), put \( x = 1 \). If you divide by \( x + 1 \), put \( x = -1 \).

📝 Teacher's Note: Show students that remainder theorem is like a shortcut. Instead of doing long division, just substitute the value and calculate. Practice with simple numbers first.

🎯 Exam Tip: Always write "By remainder theorem" first. Then clearly show f(a) calculation step by step. Check your arithmetic carefully.

Question 2. Show that:
(i) \( x - 2 \) is a factor of \( 5x^2 + 15x - 50 \)
(ii) \( 3x + 2 \) is a factor of \( 3x^2 - x - 2 \)
Answer: (x - a) is a factor of a polynomial f(x) if the remainder, when f(x) is divided by (x - a), is 0, i.e., if f(a) = 0.

(i) f(x) = \( 5x^2 + 15x - 50 \)
f(2) = \( 5(2)^2 + 15(2) - 50 = 20 + 30 - 50 = 0 \)
Hence, \( x - 2 \) is a factor of \( 5x^2 + 15x - 50 \)

(ii) f(x) = \( 3x^2 - x - 2 \)
f\( \left(-\frac{2}{3}\right) = 3\left(-\frac{2}{3}\right)^2 - \left(-\frac{2}{3}\right) - 2 = \frac{4}{3} + \frac{2}{3} - 2 = 2 - 2 = 0 \)
Hence, \( 3x + 2 \) is a factor of \( 3x^2 - x - 2 \)
In simple words: If putting a value makes the polynomial equal to zero, then that expression is a factor. It's like finding the root of the equation.

📝 Teacher's Note: Emphasize that factor means remainder is zero. Use factor theorem: if f(a) = 0, then (x - a) is a factor. Show both theorems work together.

🎯 Exam Tip: Write "By factor theorem" and clearly show f(a) = 0. For fractions like -2/3, be very careful with calculations. Show all steps.

Question 3. Use the Remainder Theorem to find which of the following is a factor of \( 2x^3 + 3x^2 - 5x - 6 \):
(i) \( x + 1 \)
(ii) \( 2x - 1 \)
(iii) \( x + 2 \)
Answer: By remainder theorem we know that when a polynomial f(x) is divided by \( x - a \), then the remainder is f(a).
Let f(x) = \( 2x^3 + 3x^2 - 5x - 6 \)

(i) f(-1) = \( 2(-1)^3 + 3(-1)^2 - 5(-1) - 6 = -2 + 3 + 5 - 6 = 0 \)
Thus, (x + 1) is a factor of the polynomial f(x).

(ii) f\( \left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 - 5\left(\frac{1}{2}\right) - 6 = \frac{1}{4} + \frac{3}{4} - \frac{5}{2} - 6 = \frac{5}{2} - 5 = \frac{-15}{2} \neq 0 \)
Thus, (2x - 1) is not a factor of the polynomial f(x).

(iii) f(-2) = \( 2(-2)^3 + 3(-2)^2 - 5(-2) - 6 = -16 + 12 + 10 - 6 = 0 \)
Thus, (x + 2) is a factor of the polynomial f(x).
In simple words: Check each option by substituting values. If the result is zero, it's a factor. If not zero, it's not a factor.

📝 Teacher's Note: Show students how to find the value to substitute. For x + 1, use x = -1. For 2x - 1, use x = 1/2. Practice finding these values first.

🎯 Exam Tip: Check all options even if you find one factor. Show complete working for each. Write "is a factor" or "is not a factor" clearly for each.

Question 4.
(i) If \( 2x + 1 \) is a factor of \( 2x^2 + ax - 3 \), find the value of a.
(ii) Find the value of k, if \( 3x - 4 \) is a factor of expression \( 3x^2 + 2x - k \).
Answer:
(i) \( 2x + 1 \) is a factor of f(x) = \( 2x^2 + ax - 3 \)
\( \therefore f\left(-\frac{1}{2}\right) = 0 \)
\( \Rightarrow 2\left(-\frac{1}{2}\right)^2 + a\left(-\frac{1}{2}\right) - 3 = 0 \)
\( \Rightarrow \frac{1}{2} - \frac{a}{2} - 3 = 0 \)
\( \Rightarrow 1 - a = 6 \)
\( \Rightarrow a = -5 \)

(ii) \( 3x - 4 \) is a factor of g(x) = \( 3x^2 + 2x - k \)
\( \therefore g\left(\frac{4}{3}\right) = 0 \)
\( \Rightarrow 3\left(\frac{4}{3}\right)^2 + 2\left(\frac{4}{3}\right) - k = 0 \)
\( \Rightarrow \frac{16}{3} + \frac{8}{3} - k = 0 \)
\( \Rightarrow \frac{24}{3} = k \)
\( \Rightarrow k = 8 \)
In simple words: When something is a factor, putting its root value makes the polynomial equal to zero. Use this fact to find unknown constants.

📝 Teacher's Note: Show students how to find the root: for 2x + 1 = 0, get x = -1/2. For 3x - 4 = 0, get x = 4/3. Then substitute these values.

🎯 Exam Tip: First find the root value correctly. Then substitute carefully and solve the equation. Double-check by putting the answer back.

Question 5. Find the values of constants a and b when \( x - 2 \) and \( x + 3 \) both are the factors of expression \( x^3 + ax^2 + bx - 12 \).
Answer: Let f(x) = \( x^3 + ax^2 + bx - 12 \)
\( x - 2 = 0 \Rightarrow x = 2 \)
\( x - 2 \) is a factor of f(x). So, remainder = 0
\( \therefore (2)^3 + a(2)^2 + b(2) - 12 = 0 \)
\( \Rightarrow 8 + 4a + 2b - 12 = 0 \)
\( \Rightarrow 4a + 2b - 4 = 0 \)
\( \Rightarrow 2a + b - 2 = 0 \) ...(1)

\( x + 3 = 0 \Rightarrow x = -3 \)
\( x + 3 \) is a factor of f(x). So, remainder = 0
\( \therefore (-3)^3 + a(-3)^2 + b(-3) - 12 = 0 \)
\( \Rightarrow -27 + 9a - 3b - 12 = 0 \)
\( \Rightarrow 9a - 3b - 39 = 0 \)
\( \Rightarrow 3a - b - 13 = 0 \) ...(2)

Adding (1) and (2), we get:
\( 5a - 15 = 0 \)
\( \Rightarrow a = 3 \)

Putting the value of a in (1), we get:
\( 6 + b - 2 = 0 \)
\( \Rightarrow b = -4 \)
In simple words: When two expressions are factors, both their root values make the polynomial zero. This gives us two equations to solve for two unknowns.

📝 Teacher's Note: Show students this creates a system of equations. One factor gives one equation, second factor gives another equation. Then solve simultaneously.

🎯 Exam Tip: Label equations as (1) and (2). Show the addition or elimination step clearly. Verify your answer by substituting both values back.

Question 6. Find the value of k, if \( 2x + 1 \) is a factor of \( (3k + 2)x^3 + (k - 1) \).
Answer: Let f(x) = \( (3k + 2)x^3 + (k - 1) \)
\( 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \)
Since, \( 2x + 1 \) is a factor of f(x), remainder is 0.
\( \therefore (3k + 2)\left(-\frac{1}{2}\right)^3 + (k - 1) = 0 \)
\( \Rightarrow -\frac{(3k + 2)}{8} + (k - 1) = 0 \)
\( \Rightarrow \frac{-3k - 2 + 8k - 8}{8} = 0 \)
\( \Rightarrow 5k - 10 = 0 \)
\( \Rightarrow k = 2 \)
In simple words: Find the root of the factor, substitute it in the polynomial, and set equal to zero to find the unknown constant.

📝 Teacher's Note: Be very careful with negative fractions and powers. Show students to work step by step with fractions. Check the arithmetic twice.

🎯 Exam Tip: Write the root clearly first. Show all fraction calculations step by step. Simplify the final equation before solving for k.

Question 7. Find the value of a, if \( x - 2 \) is a factor of \( 2x^5 - 6x^4 - 2ax^3 + 6ax^2 + 4ax + 8 \).
Answer: f(x) = \( 2x^5 - 6x^4 - 2ax^3 + 6ax^2 + 4ax + 8 \)
\( x - 2 = 0 \Rightarrow x = 2 \)
Since, \( x - 2 \) is a factor of f(x), remainder = 0.
\( 2(2)^5 - 6(2)^4 - 2a(2)^3 + 6a(2)^2 + 4a(2) + 8 = 0 \)
\( 64 - 96 - 16a + 24a + 8a + 8 = 0 \)
\( -24 + 16a = 0 \)
\( 16a = 24 \)
\( a = 1.5 \)
In simple words: Substitute x = 2 in the polynomial, collect all terms with 'a' together, and solve the simple equation to find a.

📝 Teacher's Note: Show students to substitute first, then collect like terms. Group all 'a' terms together. This makes the calculation easier and reduces errors.

🎯 Exam Tip: Calculate powers of 2 carefully: 2^5 = 32, 2^4 = 16, 2^3 = 8, 2^2 = 4. Write the final answer as a decimal or fraction as needed.

Question 8. Find the values of m and n so that \( x - 1 \) and \( x + 2 \) both are factors of \( x^3 + (3m + 1)x^2 + nx - 18 \).
Answer: Let f(x) = \( x^3 + (3m + 1)x^2 + nx - 18 \)
\( x - 1 = 0 \Rightarrow x = 1 \)
\( x - 1 \) is a factor of f(x). So, remainder = 0
\( \therefore (1)^3 + (3m + 1)(1)^2 + n(1) - 18 = 0 \)
\( \Rightarrow 1 + 3m + 1 + n - 18 = 0 \)
\( \Rightarrow 3m + n - 16 = 0 \) ...(1)

\( x + 2 = 0 \Rightarrow x = -2 \)
\( x + 2 \) is a factor of f(x). So, remainder = 0
\( \therefore (-2)^3 + (3m + 1)(-2)^2 + n(-2) - 18 = 0 \)
\( \Rightarrow -8 + 12m + 4 - 2n - 18 = 0 \)
\( \Rightarrow 12m - 2n - 22 = 0 \)
\( \Rightarrow 6m - n - 11 = 0 \) ...(2)

Adding (1) and (2), we get:
\( 9m - 27 = 0 \)
\( m = 3 \)

Putting the value of m in (1), we get:
\( 3(3) + n - 16 = 0 \)
\( 9 + n - 16 = 0 \)
\( n = 7 \)
In simple words: Two factors give two equations. Solve them together to find both unknown values. Add equations to eliminate one variable.

📝 Teacher's Note: Show students to be systematic: write both equations clearly, then use elimination method. Check the final answer by substituting back.

🎯 Exam Tip: Label equations clearly. Show elimination step. Always verify your answer by substituting both m and n values back into the original equations.

Question 9. When \( x^3 + 2x^2 - kx + 4 \) is divided by \( x - 2 \), the remainder is \( k \). Find the value of constant \( k \).
Answer:
Let \( f(x) = x^3 + 2x^2 - kx + 4 \)
\( x - 2 = 0 \Rightarrow x = 2 \)
On dividing \( f(x) \) by \( x - 2 \), it leaves a remainder \( k \).
\( \therefore f(2) = k \)
\( (2)^3 + 2(2)^2 - k(2) + 4 = k \)
\( 8 + 8 - 2k + 4 = k \)
\( 20 = 3k \)
\( k = \frac{20}{3} = 6\frac{2}{3} \)
In simple words: We use the remainder theorem. When we divide by \( (x-2) \), the remainder equals \( f(2) \). We put \( x = 2 \) in the polynomial and set it equal to \( k \).

📝 Teacher's Note: Show students that remainder theorem says remainder equals \( f(a) \) when dividing by \( (x-a) \). This makes finding remainders very easy without doing long division.

🎯 Exam Tip: Always write "By remainder theorem" first. Then substitute the value and solve. Show all steps clearly to get full marks.

Question 10. Find the value of \( a \), if the division of \( ax^3 + 9x^2 + 4x - 10 \) by \( x + 3 \) leaves a remainder 5.
Answer:
Let \( f(x) = ax^3 + 9x^2 + 4x - 10 \)
\( x + 3 = 0 \Rightarrow x = -3 \)
On dividing \( f(x) \) by \( x + 3 \), it leaves a remainder 5.
\( \therefore f(-3) = 5 \)
\( a(-3)^3 + 9(-3)^2 + 4(-3) - 10 = 5 \)
\( -27a + 81 - 12 - 10 = 5 \)
\( 54 = 27a \)
\( a = 2 \)
In simple words: We put \( x = -3 \) in the polynomial. The value we get must equal 5 (the remainder). Then we solve for \( a \).

📝 Teacher's Note: Remind students that \( x + 3 = 0 \) gives \( x = -3 \). The negative sign is very important. Many students forget this.

🎯 Exam Tip: Write "Given remainder = 5" clearly. Show the substitution step by step. Double-check your arithmetic with negative numbers.

Question 11. If \( x^3 + ax^2 + bx + 6 \) has \( x - 2 \) as a factor and leaves a remainder 3 when divided by \( x - 3 \), find the values of \( a \) and \( b \).
Answer:
Let \( f(x) = x^3 + ax^2 + bx + 6 \)
\( x - 2 = 0 \Rightarrow x = 2 \)
Since \( x - 2 \) is a factor, remainder = 0
\( \therefore f(2) = 0 \)
\( (2)^3 + a(2)^2 + b(2) + 6 = 0 \)
\( 8 + 4a + 2b + 6 = 0 \)
\( 2a + b + 7 = 0 \) ...(i)
\( x - 3 = 0 \Rightarrow x = 3 \)
On dividing \( f(x) \) by \( x - 3 \), it leaves a remainder 3.
\( \therefore f(3) = 3 \)
\( (3)^3 + a(3)^2 + b(3) + 6 = 3 \)
\( 27 + 9a + 3b + 6 = 3 \)
\( 3a + b + 10 = 0 \) ...(ii)
Subtracting (i) from (ii), we get:
\( a + 3 = 0 \)
\( a = -3 \)
Substituting the value of \( a \) in (i), we get:
\( -6 + b + 7 = 0 \)
\( b = -1 \)
In simple words: When something is a factor, the remainder is zero. When it's not a factor, we get the given remainder. We make two equations and solve them together.

📝 Teacher's Note: Teach students that "factor" means remainder is zero. Show them how to solve two equations with two unknowns by elimination or substitution method.

🎯 Exam Tip: Write "Since \( x-2 \) is a factor, \( f(2) = 0 \)" clearly. Label your equations (i) and (ii). Show the elimination step to find both values.

Question 12. The expression \( 2x^3 + ax^2 + bx - 2 \) leaves remainder 7 and 0 when divided by \( 2x - 3 \) and \( x + 2 \) respectively. Calculate the values of \( a \) and \( b \).
Answer:
Let \( f(x) = 2x^3 + ax^2 + bx - 2 \)
\( 2x - 3 = 0 \Rightarrow x = \frac{3}{2} \)
On dividing \( f(x) \) by \( 2x - 3 \), it leaves a remainder 7.
\( \therefore 2\left(\frac{3}{2}\right)^3 + a\left(\frac{3}{2}\right)^2 + b\left(\frac{3}{2}\right) - 2 = 7 \)
\( \frac{27}{4} + \frac{9a}{4} + \frac{3b}{2} = 9 \)
\( \frac{27 + 9a + 6b}{4} = 9 \)
\( 27 + 9a + 6b = 36 \)
\( 9a + 6b - 9 = 0 \)
\( 3a + 2b - 3 = 0 \) ...(i)
\( x + 2 = 0 \Rightarrow x = -2 \)
On dividing \( f(x) \) by \( x + 2 \), it leaves a remainder 0.
\( \therefore 2(-2)^3 + a(-2)^2 + b(-2) - 2 = 0 \)
\( -16 + 4a - 2b - 2 = 0 \)
\( 4a - 2b - 18 = 0 \) ...(ii)
Adding (i) and (ii), we get:
\( 7a - 21 = 0 \)
\( a = 3 \)
Substituting the value of \( a \) in (i), we get:
\( 3(3) + 2b - 3 = 0 \)
\( 9 + 2b - 3 = 0 \)
\( 2b = -6 \)
\( b = -3 \)
In simple words: We find the x-values that make each divisor zero. Then we substitute these values and set equal to the given remainders. We solve the two equations together.

📝 Teacher's Note: Show students how to handle fractions carefully when \( x = \frac{3}{2} \). Teach them to multiply through by 4 to clear fractions.

🎯 Exam Tip: Be very careful with \( 2x - 3 = 0 \) giving \( x = \frac{3}{2} \). Show all fraction arithmetic clearly. Check your final answer by substituting back.

Question 13. What number should be added to \( 3x^3 - 5x^2 + 6x \) so that when resulting polynomial is divided by \( x - 3 \), the remainder is 8?
Answer:
Let the number \( k \) be added and the resulting polynomial be \( f(x) \).
So, \( f(x) = 3x^3 - 5x^2 + 6x + k \)
It is given that when \( f(x) \) is divided by \( (x - 3) \), the remainder is 8.
\( \therefore f(3) = 8 \)
\( 3(3)^3 - 5(3)^2 + 6(3) + k = 8 \)
\( 81 - 45 + 18 + k = 8 \)
\( 54 + k = 8 \)
\( k = -46 \)
Thus, the required number is -46.
In simple words: We add an unknown number \( k \) to the polynomial. Then we use remainder theorem to find \( k \). The new polynomial at \( x = 3 \) must equal 8.

📝 Teacher's Note: Explain to students that we add \( k \) to the original polynomial. Then the remainder condition helps us find \( k \). This is a reverse problem.

🎯 Exam Tip: Write "Let the number to be added = k" first. Form the new polynomial clearly. Then apply remainder theorem to find k.

Question 14. What number should be subtracted from \( x^3 + 3x^2 - 8x + 14 \) so that on dividing it with \( x - 2 \), the remainder is 10.
Answer:
Let the number to be subtracted be \( k \) and the resulting polynomial be \( f(x) \).
So, \( f(x) = x^3 + 3x^2 - 8x + 14 - k \)
It is given that when \( f(x) \) is divided by \( (x - 2) \), the remainder is 10.
\( \therefore f(2) = 10 \)
\( (2)^3 + 3(2)^2 - 8(2) + 14 - k = 10 \)
\( 8 + 12 - 16 + 14 - k = 10 \)
\( 18 - k = 10 \)
\( k = 8 \)
Thus, the required number is 8.
In simple words: We subtract an unknown number \( k \) from the polynomial. Then we use remainder theorem to find \( k \). The new polynomial at \( x = 2 \) must equal 10.

📝 Teacher's Note: Show the difference between adding and subtracting. When we subtract \( k \), the new polynomial becomes original minus \( k \).

🎯 Exam Tip: Write "Let the number to be subtracted = k" first. The new polynomial is original minus k. Apply remainder theorem correctly.

Question 15. The polynomials \( 2x^3 - 7x^2 + ax - 6 \) and \( x^3 - 8x^2 + (2a + 1)x - 16 \) leaves the same remainder when divided by \( x - 2 \). Find the value of 'a'.
Answer:
Let \( f(x) = 2x^3 - 7x^2 + ax - 6 \)
\( x - 2 = 0 \Rightarrow x = 2 \)
When \( f(x) \) is divided by \( (x - 2) \), remainder = \( f(2) \)
\( \therefore f(2) = 2(2)^3 - 7(2)^2 + a(2) - 6 \)
\( = 16 - 28 + 2a - 6 \)
\( = 2a - 18 \)
Let \( g(x) = x^3 - 8x^2 + (2a + 1)x - 16 \)
When \( g(x) \) is divided by \( (x - 2) \), remainder = \( g(2) \)
\( \therefore g(2) = (2)^3 - 8(2)^2 + (2a + 1)(2) - 16 \)
\( = 8 - 32 + 4a + 2 - 16 \)
\( = 4a - 38 \)
By the given condition, we have:
\( f(2) = g(2) \)
\( 2a - 18 = 4a - 38 \)
\( 4a - 2a = 38 - 18 \)
\( 2a = 20 \)
\( a = 10 \)
Thus, the value of a is 10.
In simple words: Both polynomials give the same remainder when divided by \( (x-2) \). So we find both remainders using remainder theorem and set them equal.

📝 Teacher's Note: Show students that "same remainder" means \( f(2) = g(2) \). This gives us one equation in one unknown, which is easy to solve.

🎯 Exam Tip: Write both remainders separately first. Then write "By given condition, \( f(2) = g(2) \)". Show the algebra steps clearly to find a.

Question 16. If \( (x - 2) \) is a factor of the expression \( 2x^3 + ax^2 + bx - 14 \) and when the expression is divided by \( (x - 3) \), it leaves a remainder 52, find the values of a and b.
Answer:
Since \( (x - 2) \) is a factor of polynomial \( 2x^3 + ax^2 + bx - 14 \), we have
\( 2(2)^3 + a(2)^2 + b(2) - 14 = 0 \)
\( \Rightarrow 16 + 4a + 2b - 14 = 0 \)
\( \Rightarrow 4a + 2b + 2 = 0 \)
\( \Rightarrow 2a + b + 1 = 0 \)
\( \Rightarrow 2a + b = -1 \) ....(i)
On dividing by \( (x - 3) \), the polynomial \( 2x^3 + ax^2 + bx - 14 \) leaves remainder 52,
\( \Rightarrow 2(3)^3 + a(3)^2 + b(3) - 14 = 52 \)
\( \Rightarrow 54 + 9a + 3b - 14 = 52 \)
\( \Rightarrow 9a + 3b + 40 = 52 \)
\( \Rightarrow 9a + 3b = 12 \)
\( \Rightarrow 3a + b = 4 \) ....(ii)
Subtracting (i) from (ii), we get
\( a = 5 \)
Substituting \( a = 5 \) in (i), we get
\( 2 \times 5 + b = -1 \)
\( \Rightarrow 10 + b = -1 \)
\( \Rightarrow b = -11 \)
Hence, \( a = 5 \) and \( b = -11 \).
In simple words: Factor means remainder is zero. Non-factor gives the specified remainder. We use both conditions to make two equations and solve them.

📝 Teacher's Note: Emphasize that factor means the polynomial equals zero at that point. Non-factor gives the stated remainder value.

🎯 Exam Tip: Write "Since (x-2) is a factor, remainder = 0" clearly. Set up both equations properly and solve by elimination method.

Question 17. Find 'a' if the two polynomials \( ax^3 + 3x^2 - 9 \) and \( 2x^3 + 4x + a \) leave the same remainder when divided by \( x + 3 \).
Answer:
\( x + 3 = 0 \Rightarrow x = -3 \)
Since the given polynomials leave the same remainder when divided by \( (x + 3) \),
Value of polynomial \( ax^3 + 3x^2 - 9 \) at \( x = -3 \) is same as value of polynomial \( 2x^3 + 4x + a \) at \( x = -3 \).
\( a(-3)^3 + 3(-3)^2 - 9 = 2(-3)^3 + 4(-3) + a \)
\( -27a + 27 - 9 = -54 - 12 + a \)
\( -27a + 18 = -66 + a \)
\( -27a - a = -66 - 18 \)
\( -28a = -84 \)
\( a = 3 \)
In simple words: Both polynomials give the same value when we put \( x = -3 \). So we substitute \( x = -3 \) in both and set them equal.

📝 Teacher's Note: Show students that same remainder means the function values are equal at the root of the divisor. This creates one equation to solve.

🎯 Exam Tip: Write the equation clearly: "Value of first polynomial at x = -3 equals value of second polynomial at x = -3". Solve the algebra step by step.

Exercise 8B

Question 1. Using the Factor Theorem, show that:
(i) (x – 2) is a factor of \( x^3 - 2x^2 - 9x + 18 \). Hence, factorise the expression \( x^3 - 2x^2 - 9x + 18 \) completely.
(ii) (x + 5) is a factor of \( 2x^3 + 5x^2 - 28x - 15 \). Hence, factorise the expression \( 2x^3 + 5x^2 - 28x - 15 \) completely.
(iii) (3x + 2) is a factor of \( 3x^3 + 2x^2 - 3x - 2 \). Hence, factorise the expression \( 3x^3 + 2x^2 - 3x - 2 \) completely.
Answer:
(i) Let f(x) = \( x^3 - 2x^2 - 9x + 18 \)
x - 2 = 0
\( \implies \) x = 2

Remainder = f(2)
= \( (2)^3 - 2(2)^2 - 9(2) + 18 \)
= 8 - 8 - 18 + 18
= 0
Hence, (x - 2) is a factor of f(x).

Now, we have polynomial division:
Using long division: \( x^3 - 2x^2 - 9x + 18 = (x - 2)(x^2 - 9) \)
\( = (x - 2)(x + 3)(x - 3) \)

(ii) Let f(x) = \( 2x^3 + 5x^2 - 28x - 15 \)
x + 5 = 0
\( \implies \) x = -5

Remainder = f(-5)
= \( 2(-5)^3 + 5(-5)^2 - 28(-5) - 15 \)
= -250 + 125 + 140 - 15
= -265 + 265
= 0
Hence, (x + 5) is a factor of f(x).

Using polynomial division:
\( 2x^3 + 5x^2 - 28x - 15 = (x + 5)(2x^2 - 5x - 3) \)
= (x + 5)[2x(x - 3) + 1(x - 3)]
= (x + 5)(2x + 1)(x - 3)

(iii) Let f(x) = \( 3x^3 + 2x^2 - 3x - 2 \)
3x + 2 = 0
\( \implies \) x = \( -\frac{2}{3} \)

Remainder = f(\( -\frac{2}{3} \))
= \( 3(-\frac{2}{3})^3 + 2(-\frac{2}{3})^2 - 3(-\frac{2}{3}) - 2 \)
= \( -\frac{8}{9} + \frac{8}{9} + 2 - 2 \)
= 0
Hence, (3x + 2) is a factor of f(x).

Using polynomial division:
\( 3x^3 + 2x^2 - 3x - 2 = (3x + 2)(x^2 - 1) = (3x + 2)(x + 1)(x - 1) \)

In simple words: Factor Theorem says if we put a value of x in a polynomial and get 0, then (x - that value) is a factor. We use division to find the other factors.

📝 Teacher's Note: Show students that Factor Theorem is like a test. If f(a) = 0, then (x - a) is a factor. Always check by substituting the value first.

🎯 Exam Tip: First prove the given factor by showing remainder is zero. Then do polynomial division to find complete factorization. Write each step clearly.

Question 2. Using the Remainder Theorem, factorise each of the following completely.
(i) \( 3x^3 + 2x^2 - 19x + 6 \)
(ii) \( 2x^3 + x^2 - 13x + 6 \)
(iii) \( 3x^3 + 2x^2 - 23x - 30 \)
(iv) \( 4x^3 + 7x^2 - 36x - 63 \)
(v) \( x^3 + x^2 - 4x - 4 \)
Answer:
(i) For \( 3x^3 + 2x^2 - 19x + 6 \)
For x = 2, the value of the given expression:
\( 3(2)^3 + 2(2)^2 - 19(2) + 6 \)
= 24 + 8 - 38 + 6
= 0

\( \implies \) x - 2 is a factor of \( 3x^3 + 2x^2 - 19x + 6 \)

Using long division:
\( 3x^3 + 2x^2 - 19x + 6 = (x - 2)(3x^2 + 8x - 3) \)
= (x - 2)(3x(x + 3) - (x + 3))
= (x - 2)(3x - 1)(x + 3)

(ii) Let f(x) = \( 2x^3 + x^2 - 13x + 6 \)
For x = 2,
f(x) = f(2) = \( 2(2)^3 + (2)^2 - 13(2) + 6 = 16 + 4 - 26 + 6 = 0 \)
Hence, (x - 2) is a factor of f(x).

Using long division:
\( 2x^3 + x^2 - 13x + 6 = (x - 2)(2x^2 + 5x - 3) \)
= (x - 2)(2x^2 + 6x - x - 3)
= (x - 2)[2x(x + 3) - (x + 3)]
= (x - 2)(x + 3)(2x - 1)

(iii) f(x) = \( 3x^3 + 2x^2 - 23x - 30 \)
For x = -2,
f(x) = f(-2) = \( 3(-2)^3 + 2(-2)^2 - 23(-2) - 30 \)
= -24 + 8 + 46 - 30 = -54 + 54 = 0
Hence, (x + 2) is a factor of f(x).

Using long division:
\( 3x^3 + 2x^2 - 23x - 30 = (x + 2)(3x^2 - 4x - 15) \)
= (x + 2)(3x^2 + 5x - 9x - 15)
= (x + 2)[x(3x + 5) - 3(3x + 5)]
= (x + 2)(3x + 5)(x - 3)

(iv) f(x) = \( 4x^3 + 7x^2 - 36x - 63 \)
For x = 3,
f(x) = f(3) = \( 4(3)^3 + 7(3)^2 - 36(3) - 63 \)
= 108 + 63 - 108 - 63 = 0
Hence, (x + 3) is a factor of f(x).

Using long division:
\( 4x^3 + 7x^2 - 36x - 63 = (x + 3)(4x^2 - 5x - 21) \)
= (x + 3)(\( 4x^2 - 12x + 7x - 21 \))
= (x + 3)[4x(x - 3) + 7(x - 3)]
= (x + 3)(4x + 7)(x - 3)

(v) f(x) = \( x^3 + x^2 - 4x - 4 \)
For x = 2,
f(x) = f(2) = \( (2)^3 + (2)^2 - 4(2) - 4 = 8 + 4 - 8 - 4 = 0 \)
Hence, (x - 2) is a factor of f(x).

Using long division:
\( x^3 + x^2 - 4x - 4 = (x - 2)(x^2 + 3x + 2) \)
= (x - 2)(x^2 + 2x + x + 2)
= (x - 2)[x(x + 2) + 1(x + 2)]
= (x - 2)(x + 1)(x + 2)

In simple words: We test different values like 1, 2, -1, -2, etc. in the polynomial. When we get 0, that tells us we found a factor. Then we divide to get the remaining factors.

📝 Teacher's Note: Students should try simple values like ±1, ±2, ±3 first. Make them understand that finding one factor makes the rest easier by division.

🎯 Exam Tip: Always show the substitution step where f(a) = 0. Then write the complete factorization clearly. Check your answer by expanding back if time allows.

Question 3. Using the Remainder Theorem, factorise the expression \(3x^3 + 10x^2 + x - 6\). Hence, solve the equation \(3x^3 + 10x^2 + x - 6 = 0\).
Answer:
Let \(f(x) = 3x^3 + 10x^2 + x - 6\)
For \(x = -1\),
\(f(x) = f(-1) = 3(-1)^3 + 10(-1)^2 + (-1) - 6 = -3 + 10 - 1 - 6 = 0\)
Hence, \((x + 1)\) is a factor of \(f(x)\).

Using polynomial division:
\(3x^3 + 10x^2 + x - 6 = (x + 1)(3x^2 + 7x - 6)\)
\[= (x + 1)(3x^2 + 9x - 2x - 6)\]
\[= (x + 1)[3x(x + 3) - 2(x + 3)]\]
\[= (x + 1)(x + 3)(3x - 2)\]

Now, \(3x^3 + 10x^2 + x - 6 = 0\)
\[\implies (x + 1)(x + 3)(3x - 2) = 0\]
\[\implies x = -1, -3, \frac{2}{3}\]

In simple words: We test values to find one factor. Then we divide to get the other factors. The equation has three solutions where each factor equals zero.

📝 Teacher's Note: First test simple values like -1, 1, 2 in the Remainder Theorem. Once you find one factor, use long division to find the rest. Show each step clearly.

🎯 Exam Tip: Always verify your factors by multiplying them back. Write the final answer as \(x = -1, -3, \frac{2}{3}\) to get full marks.

Question 4. Factorise the expression \(f(x) = 2x^3 - 7x^2 - 3x + 18\). Hence, find all possible values of x for which \(f(x) = 0\).
Answer:
Let \(f(x) = 2x^3 - 7x^2 - 3x + 18\)
For \(x = 2\),
\(f(x) = f(2) = 2(2)^3 - 7(2)^2 - 3(2) + 18\)
\[= 16 - 28 - 6 + 18 = 0\]
Hence, \((x - 2)\) is a factor of \(f(x)\).

Using polynomial division:
\(2x^3 - 7x^2 - 3x + 18 = (x - 2)(2x^2 - 3x - 9)\)
\[= (x - 2)(2x^2 - 6x + 3x - 9)\]
\[= (x - 2)[2x(x - 3) + 3(x - 3)]\]
\[= (x - 2)(x - 3)(2x + 3)\]

Now, \(f(x) = 0\)
\[\implies 2x^3 - 7x^2 - 3x + 18 = 0\]
\[\implies (x - 2)(x - 3)(2x + 3) = 0\]
\[\implies x = 2, 3, -\frac{3}{2}\]

In simple words: We test values to find one factor. Then we divide to get all factors. The solutions are the values that make each factor zero.

📝 Teacher's Note: Try factors of the constant term divided by factors of the leading coefficient. For this problem, try ±1, ±2, ±3, ±9, ±½, ±3/2, ±9/2.

🎯 Exam Tip: Write the complete factorization first, then solve each factor equals zero. List all solutions clearly: \(x = 2, 3, -\frac{3}{2}\).

Question 5. Given that \(x - 2\) and \(x + 1\) are factors of \(f(x) = x^3 + 3x^2 + ax + b\); calculate the values of a and b. Hence, find all the factors of \(f(x)\).
Answer:
\(f(x) = x^3 + 3x^2 + ax + b\)

Since \((x - 2)\) is a factor of \(f(x)\), \(f(2) = 0\)
\[\implies (2)^3 + 3(2)^2 + a(2) + b = 0\]
\[\implies 8 + 12 + 2a + b = 0\]
\[\implies 2a + b + 20 = 0 ...(i)\]

Since \((x + 1)\) is a factor of \(f(x)\), \(f(-1) = 0\)
\[\implies (-1)^3 + 3(-1)^2 + a(-1) + b = 0\]
\[\implies -1 + 3 - a + b = 0\]
\[\implies -a + b + 2 = 0 ...(ii)\]

Subtracting (ii) from (i), we get:
\(3a + 18 = 0\)
\[\implies a = -6\]

Substituting the value of a in (ii), we get:
\(b = a - 2 = -6 - 2 = -8\)

\[\therefore f(x) = x^3 + 3x^2 - 6x - 8\]

Now, for \(x = -1\),
\(f(x) = f(-1) = (-1)^3 + 3(-1)^2 - 6(-1) - 8 = -1 + 3 + 6 - 8 = 0\)
Hence, \((x + 1)\) is a factor of \(f(x)\).

Using polynomial division:
\(x^3 + 3x^2 - 6x - 8 = (x + 1)(x^2 + 2x - 8)\)
\[= (x + 1)(x^2 + 4x - 2x - 8)\]
\[= (x + 1)[x(x + 4) - 2(x + 4)]\]
\[= (x + 1)(x + 4)(x - 2)\]

In simple words: We use the fact that if a value is a factor, the function equals zero at that value. This gives us two equations to find a and b. Then we can find all factors.

📝 Teacher's Note: When given factors, substitute the roots into the function and set equal to zero. This creates a system of equations to solve for unknown coefficients.

🎯 Exam Tip: Write \(a = -6\) and \(b = -8\) clearly. Then write the complete factorization: \(f(x) = (x + 1)(x + 4)(x - 2)\).

Question 6. The expression \(4x^3 - bx^2 + x - c\) leaves remainders 0 and 30 when divided by \(x + 1\) and \(2x - 3\) respectively. Calculate the values of b and c. Hence, factorise the expression completely.
Answer:
Let \(f(x) = 4x^3 - bx^2 + x - c\)

It is given that when \(f(x)\) is divided by \((x + 1)\), the remainder is 0.
\[\therefore f(-1) = 0\]
\[4(-1)^3 - b(-1)^2 + (-1) - c = 0\]
\[-4 - b - 1 - c = 0\]
\[b + c + 5 = 0 ...(i)\]

It is given that when \(f(x)\) is divided by \((2x - 3)\), the remainder is 30.
\[\therefore f\left(\frac{3}{2}\right) = 30\]
\[4\left(\frac{3}{2}\right)^3 - b\left(\frac{3}{2}\right)^2 + \left(\frac{3}{2}\right) - c = 30\]
\[\frac{27}{2} - \frac{9b}{4} + \frac{3}{2} - c = 30\]
\[54 - 9b + 6 - 4c = 120\]
\[9b + 4c + 60 = 0 ...(ii)\]

Multiplying (i) by 4 and subtracting it from (ii), we get:
\[5b + 40 = 0\]
\[b = -8\]

Substituting the value of b in (i), we get:
\[c = -5 + 8 = 3\]

Therefore, \(f(x) = 4x^3 + 8x^2 + x - 3\)

Now, for \(x = -1\), we get:
\[f(x) = f(-1) = 4(-1)^3 + 8(-1)^2 + (-1) - 3 = -4 + 8 - 1 - 3 = 0\]

Hence, \((x + 1)\) is a factor of \(f(x)\).

Using polynomial division:
\(4x^3 + 8x^2 + x - 3 = (x + 1)(4x^2 + 4x - 3)\)
\[= (x + 1)(4x^2 + 6x - 2x - 3)\]
\[= (x + 1)[2x(2x + 3) - 1(2x + 3)]\]
\[= (x + 1)(2x + 3)(2x - 1)\]

In simple words: We use the remainder theorem. When divided by a linear factor, the remainder equals the function value at the root. This gives us two equations to solve.

📝 Teacher's Note: For \(2x - 3\), the root is \(\frac{3}{2}\), not \(-3\). Students often make this mistake. Remember to solve the linear factor equals zero first.

🎯 Exam Tip: Write \(b = -8\) and \(c = 3\) clearly. Then show the complete factorization: \((x + 1)(2x + 3)(2x - 1)\).

Question 7. If \(x + a\) is a common factor of expressions \(f(x) = x^2 + px + q\) and \(g(x) = x^2 + mx + n\); show that: \(a = \frac{n - q}{m - p}\).
Answer:
\(f(x) = x^2 + px + q\)
It is given that \((x + a)\) is a factor of \(f(x)\).
\[\therefore f(-a) = 0\]
\[\implies (-a)^2 + p(-a) + q = 0\]
\[\implies a^2 - pa + q = 0\]
\[\implies a^2 = pa - q ...(i)\]

\(g(x) = x^2 + mx + n\)
It is given that \((x + a)\) is a factor of \(g(x)\).
\[\therefore g(-a) = 0\]
\[\implies (-a)^2 + m(-a) + n = 0\]
\[\implies a^2 - ma + n = 0\]
\[\implies a^2 = ma - n ...(ii)\]

From (i) and (ii), we get:
\[pa - q = ma - n\]
\[pa - ma = q - n\]
\[a(p - m) = q - n\]
\[a = \frac{q - n}{p - m} = \frac{n - q}{m - p}\]

In simple words: If a linear expression is a common factor, both functions equal zero at the same root. This gives us two equations with the same \(a^2\) value, which we can equate.

📝 Teacher's Note: Emphasize that a common factor means both functions have the same root. Students should remember that if \((x + a)\) is a factor, then substituting \(x = -a\) gives zero.

🎯 Exam Tip: Show both equations \(a^2 = pa - q\) and \(a^2 = ma - n\) clearly. Then equate them to get the final result. The algebra must be shown step by step.

Question 8. The polynomials ax³ + 3x² – 3 and 2x³ – 5x + a, when divided by x – 4, leave the same remainder in each case. Find the value of a.
Answer:
Step 1: Find remainder when first polynomial is divided by (x – 4).
Let \( f(x) = ax^3 + 3x^2 - 3 \)
When f(x) is divided by (x – 4), remainder = f(4)
\( f(4) = a(4)^3 + 3(4)^2 - 3 = 64a + 45 \)

Step 2: Find remainder when second polynomial is divided by (x – 4).
Let \( g(x) = 2x^3 - 5x + a \)
When g(x) is divided by (x – 4), remainder = g(4)
\( g(4) = 2(4)^3 - 5(4) + a = a + 108 \)

Step 3: Set the remainders equal.
It is given that f(4) = g(4)
\( 64a + 45 = a + 108 \)
\( 63a = 63 \)
\( a = 1 \)
In simple words: We use the remainder theorem. When any polynomial is divided by (x – 4), the remainder is what we get when we put x = 4 in the polynomial. Since both remainders are equal, we make them equal and solve for a.

📝 Teacher's Note: Show students that remainder theorem says: when f(x) is divided by (x – a), remainder = f(a). This makes finding remainders very easy without long division.

🎯 Exam Tip: Always write "By remainder theorem, remainder = f(a)" first. Then substitute x = a in both polynomials. Set them equal and solve.

Question 9. Find the value of 'a', if (x – a) is a factor of x³ – ax² + x + 2.
Answer:
Let \( f(x) = x^3 - ax^2 + x + 2 \)
It is given that (x – a) is a factor of f(x).
By factor theorem, remainder = f(a) = 0
\( a^3 - a \cdot a^2 + a + 2 = 0 \)
\( a^3 - a^3 + a + 2 = 0 \)
\( a + 2 = 0 \)
\( a = -2 \)
In simple words: If (x – a) is a factor, then when we put x = a in the polynomial, we get zero. We substitute x = a and solve the equation to find a.

📝 Teacher's Note: Remind students that factor theorem is the reverse of remainder theorem. If remainder is zero, then the divisor is a factor.

🎯 Exam Tip: Write "By factor theorem, f(a) = 0" first. Then substitute x = a carefully. Many students make mistakes in the substitution step.

Question 10. Find the number that must be subtracted from the polynomial 3y³ + y² – 22y + 15, so that the resulting polynomial is completely divisible by y + 3.
Answer:
Let the number to be subtracted from the given polynomial be k.
Let \( f(y) = 3y^3 + y^2 - 22y + 15 - k \)
It is given that f(y) is divisible by (y + 3).
By factor theorem, remainder = f(-3) = 0
\( 3(-3)^3 + (-3)^2 - 22(-3) + 15 - k = 0 \)
\( -81 + 9 + 66 + 15 - k = 0 \)
\( 9 - k = 0 \)
\( k = 9 \)
In simple words: We need to subtract a number so that when we divide by (y + 3), we get remainder zero. We find this number by using the factor theorem.

📝 Teacher's Note: Explain that subtracting k means the new polynomial is the old one minus k. Then use factor theorem with the divisor y + 3.

🎯 Exam Tip: Let the number be k. Write the new polynomial as original minus k. Then use f(-3) = 0 to find k. Don't forget the negative sign.

Exercise 8C

Question 1. Show that (x – 1) is a factor of x³ – 7x² + 14x – 8. Hence, completely factorise the given expression.
Answer:
Let \( f(x) = x^3 - 7x^2 + 14x - 8 \)
\( f(1) = (1)^3 - 7(1)^2 + 14(1) - 8 = 1 - 7 + 14 - 8 = 0 \)
Hence, (x – 1) is a factor of f(x).

Now dividing \( x^3 - 7x^2 + 14x - 8 \) by (x – 1):
\( x^3 - 7x^2 + 14x - 8 = (x - 1)(x^2 - 6x + 8) \)

Factoring \( x^2 - 6x + 8 \):
\( x^2 - 6x + 8 = (x - 2)(x - 4) \)

Therefore: \( x^3 - 7x^2 + 14x - 8 = (x - 1)(x - 2)(x - 4) \)
In simple words: First we check if (x – 1) is a factor by putting x = 1. Then we divide to get a quadratic. Finally we factor the quadratic to get all three factors.

📝 Teacher's Note: Show the long division step by step on the board. Students often make errors in polynomial division. Practice this method multiple times.

🎯 Exam Tip: First prove it's a factor by showing f(1) = 0. Then do polynomial division carefully. Finally factor the quotient if possible. Show all steps.

Question 2. Using Remainder Theorem, factorise: x³ + 10x² – 37x + 26 completely.
Answer:
By Remainder Theorem,
For x = 1, the value of the given expression is the remainder.
\( x^3 + 10x^2 - 37x + 26 \)
\( = (1)^3 + 10(1)^2 - 37(1) + 26 \)
\( = 1 + 10 - 37 + 26 \)
\( = 37 - 37 \)
\( = 0 \)

Therefore, (x – 1) is a factor of \( x^3 + 10x^2 - 37x + 26 \).

Dividing by (x – 1):
\( x^3 + 10x^2 - 37x + 26 = (x - 1)(x^2 + 11x - 26) \)

Factoring \( x^2 + 11x - 26 \):
\( x^2 + 11x - 26 = (x + 13)(x - 2) \)

Therefore: \( x^3 + 10x^2 - 37x + 26 = (x - 1)(x + 13)(x - 2) \)
In simple words: We try small values like x = 1, x = -1, x = 2 etc. in the polynomial. When we get zero, that tells us we found a factor. Then we divide and repeat.

📝 Teacher's Note: Teach students to try simple values first: ±1, ±2, ±factors of constant term. This saves time in finding the first factor.

🎯 Exam Tip: Always try x = 1 first, then x = -1, then x = 2, etc. Once you find one factor, use polynomial division to find the rest.

Question 3. When x³ + 3x² – mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.
Answer:
Let \( f(x) = x^3 + 3x^2 - mx + 4 \)
According to the given information,
f(2) = m + 3
\( (2)^3 + 3(2)^2 - m(2) + 4 = m + 3 \)
\( 8 + 12 - 2m + 4 = m + 3 \)
\( 24 - 3 = m + 2m \)
\( 3m = 21 \)
\( m = 7 \)
In simple words: We use remainder theorem. The remainder when dividing by (x – 2) is f(2). We are told this remainder equals m + 3, so we solve f(2) = m + 3.

📝 Teacher's Note: Make sure students understand that the remainder is given in the problem. We don't need to calculate it separately - just use the remainder theorem formula.

🎯 Exam Tip: Write "f(2) = m + 3" as the first step. Substitute x = 2 on the left side, keep m + 3 on the right side. Then solve for m.

Question 4. What should be subtracted from 3x³ – 8x² + 4x – 3, so that the resulting expression has x + 2 as a factor?
Answer:
Let the required number be k.
Let \( f(x) = 3x^3 - 8x^2 + 4x - 3 - k \)
According to the given information,
f(-2) = 0
\( 3(-2)^3 - 8(-2)^2 + 4(-2) - 3 - k = 0 \)
\( -24 - 32 - 8 - 3 - k = 0 \)
\( -67 - k = 0 \)
\( k = -67 \)
Thus, the required number is -67.
In simple words: We subtract k from the polynomial. For (x + 2) to be a factor, the new polynomial must equal zero when x = -2. We solve to find k.

📝 Teacher's Note: Remind students that subtracting a negative number means adding a positive number. So we actually need to add 67 to the original polynomial.

🎯 Exam Tip: Let the number to be subtracted be k. Write new polynomial as original minus k. Use f(-2) = 0 since we want (x + 2) as factor.

Question 5. If (x + 1) and (x – 2) are factors of x³ + (a + 1)x² – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.
Answer:
Let \( f(x) = x^3 + (a + 1)x^2 - (b - 2)x - 6 \)

Since (x + 1) is a factor of f(x):
Remainder = f(-1) = 0
\( (-1)^3 + (a + 1)(-1)^2 - (b - 2)(-1) - 6 = 0 \)
\( -1 + (a + 1) + (b - 2) - 6 = 0 \)
\( a + b - 8 = 0 \) ...(i)

Since (x – 2) is a factor of f(x):
Remainder = f(2) = 0
\( (2)^3 + (a + 1)(2)^2 - (b - 2)(2) - 6 = 0 \)
\( 8 + 4a + 4 - 2b + 4 - 6 = 0 \)
\( 4a - 2b + 10 = 0 \)
\( 2a - b + 5 = 0 \) ...(ii)

Adding (i) and (ii):
\( 3a - 3 = 0 \)
\( a = 1 \)

Substituting a = 1 in (i):
\( 1 + b - 8 = 0 \)
\( b = 7 \)

So \( f(x) = x^3 + 2x^2 - 5x - 6 \)
Since (x + 1) and (x – 2) are factors, \( (x + 1)(x - 2) = x^2 - x - 2 \) is a factor.
Dividing: \( f(x) = x^3 + 2x^2 - 5x - 6 = (x + 1)(x - 2)(x + 3) \)
In simple words: We use the fact that both (x + 1) and (x – 2) are factors. This gives us two equations with f(-1) = 0 and f(2) = 0. We solve these to find a and b.

📝 Teacher's Note: Show students how to solve the system of equations step by step. Adding equations to eliminate variables is a useful technique to practice.

🎯 Exam Tip: Use f(-1) = 0 for factor (x + 1) and f(2) = 0 for factor (x – 2). Solve the system carefully. Then substitute back to get the complete factorization.

Question 6. If x – 2 is a factor of x² + ax + b and a + b = 1, find the values of a and b.
Answer:
Let \( f(x) = x^2 + ax + b \)
Since (x – 2) is a factor of f(x):
Remainder = f(2) = 0
\( (2)^2 + a(2) + b = 0 \)
\( 4 + 2a + b = 0 \)
\( 2a + b = -4 \) ...(i)

It is given that:
\( a + b = 1 \) ...(ii)

Subtracting (ii) from (i):
\( a = -5 \)

Substituting a = -5 in (ii):
\( b = 1 - (-5) = 6 \)
In simple words: We have two conditions: (x – 2) is a factor and a + b = 1. These give us two equations. We solve them together to find a and b.

📝 Teacher's Note: This is a good example of using two different conditions to create a system of equations. Both pieces of information are equally important.

🎯 Exam Tip: Use f(2) = 0 for the factor condition. Write both equations clearly. Subtract one from the other to eliminate b and find a first.

Question 7. Factorise x³ + 6x² + 11x + 6 completely using factor theorem.
Answer:
Let \( f(x) = x^3 + 6x^2 + 11x + 6 \)
For x = -1:
\( f(-1) = (-1)^3 + 6(-1)^2 + 11(-1) + 6 \)
\( = -1 + 6 - 11 + 6 = 12 - 12 = 0 \)
Hence, (x + 1) is a factor of f(x).

Dividing by (x + 1), we get:
\( f(x) = (x + 1)(x^2 + 5x + 6) \)

Factoring \( x^2 + 5x + 6 \):
\( x^2 + 5x + 6 = (x + 2)(x + 3) \)

Therefore: \( x^3 + 6x^2 + 11x + 6 = (x + 1)(x + 2)(x + 3) \)
In simple words: We try x = -1 and get zero, so (x + 1) is a factor. We divide to get a quadratic, then factor that quadratic to get all three factors.

📝 Teacher's Note: Since all coefficients are positive and the constant term is positive, try negative values first. This often works for such polynomials.

🎯 Exam Tip: Try x = -1, -2, -3 etc. when all terms are positive. Once you find one factor, divide and factor the quotient. Write the complete factorization.

Question 8. Find the value of 'm', if \( mx^3 + 2x^2 - 3 \) and \( x^2 - mx + 4 \) leave the same remainder when each is divided by \( x - 2 \).
Answer:
Given:
Let \( f(x) = mx^3 + 2x^2 - 3 \)
Let \( g(x) = x^2 - mx + 4 \)

It is given that \( f(x) \) and \( g(x) \) leave the same remainder when divided by \( (x - 2) \).

Step 1: Use the Remainder Theorem.
For same remainders: \( f(2) = g(2) \)

Step 2: Calculate \( f(2) \).
\( f(2) = m(2)^3 + 2(2)^2 - 3 = 8m + 8 - 3 = 8m + 5 \)

Step 3: Calculate \( g(2) \).
\( g(2) = (2)^2 - m(2) + 4 = 4 - 2m + 4 = 8 - 2m \)

Step 4: Set \( f(2) = g(2) \).
\( 8m + 5 = 8 - 2m \)
\( 8m + 2m = 8 - 5 \)
\( 10m = 3 \)
\( m = \frac{3}{10} \)

Answer: \( m = \frac{3}{10} \)
In simple words: When two polynomials give the same remainder, we use the Remainder Theorem. We put the same value in both and make them equal.

📝 Teacher's Note: Teach students that remainder when dividing by \( (x - a) \) is found by putting \( x = a \) in the polynomial. This is the Remainder Theorem.

🎯 Exam Tip: Always write "Using Remainder Theorem" at the start. Show all steps clearly. Don't forget to simplify the fraction answer.

Question 9. The polynomial \( px^3 + 4x^2 - 3x + q \) is completely divisible by \( x^2 - 1 \); find the values of p and q. Also, for these values of p and q factorize the given polynomial completely.
Answer:
Given:
Let \( f(x) = px^3 + 4x^2 - 3x + q \)
It is given that \( f(x) \) is completely divisible by \( (x^2 - 1) = (x + 1)(x - 1) \).

Step 1: For complete divisibility, both factors must divide the polynomial.
Therefore, \( f(1) = 0 \) and \( f(-1) = 0 \)

Step 2: Calculate \( f(1) = 0 \).
\( f(1) = p(1)^3 + 4(1)^2 - 3(1) + q = 0 \)
\( p + 4 - 3 + q = 0 \)
\( p + q + 1 = 0 \) ...(i)

Step 3: Calculate \( f(-1) = 0 \).
\( f(-1) = p(-1)^3 + 4(-1)^2 - 3(-1) + q = 0 \)
\( -p + 4 + 3 + q = 0 \)
\( -p + q + 7 = 0 \) ...(ii)

Step 4: Solve the system of equations.
Adding (i) and (ii): \( 2q + 8 = 0 \)
\( q = -4 \)

Substituting in (i): \( p = -q - 1 = -(-4) - 1 = 4 - 1 = 3 \)

Step 5: The polynomial becomes \( f(x) = 3x^3 + 4x^2 - 3x - 4 \)

Step 6: Complete factorization by division.
[Diagram: Long division showing \( 3x^3 + 4x^2 - 3x - 4 = (x^2 - 1)(3x + 4) \)]

Therefore: \( 3x^3 + 4x^2 - 3x - 4 = (x^2 - 1)(3x + 4) = (x - 1)(x + 1)(3x + 4) \)

Answer: p = 3, q = -4
Factorization: \( (x - 1)(x + 1)(3x + 4) \)
In simple words: When a polynomial is divisible by \( (x^2 - 1) \), it must be divisible by both \( (x - 1) \) and \( (x + 1) \). So we put both values and get two equations.

📝 Teacher's Note: Show students that \( x^2 - 1 = (x - 1)(x + 1) \) first. Then explain that if both factors divide, both roots make the polynomial zero.

🎯 Exam Tip: Always write both conditions \( f(1) = 0 \) and \( f(-1) = 0 \). Show the long division step to get full marks for factorization.

Question 10. Find the number which should be added to \( x^2 + x + 3 \) so that the resulting polynomial is completely divisible by \( (x + 3) \).
Answer:
Given:
Let the required number be k.
Let \( f(x) = x^2 + x + 3 + k \)

Step 1: For complete divisibility by \( (x + 3) \).
Remainder = 0
Therefore, \( f(-3) = 0 \)

Step 2: Calculate \( f(-3) \).
\( f(-3) = (-3)^2 + (-3) + 3 + k = 0 \)
\( 9 - 3 + 3 + k = 0 \)
\( 9 + k = 0 \)
\( k = -9 \)

Answer: The required number is -9.
In simple words: We need to add a number so that when we put \( x = -3 \), the polynomial becomes zero. That number is -9.

📝 Teacher's Note: Remind students that divisible by \( (x + 3) \) means the remainder is zero when \( x = -3 \). Be careful with signs.

🎯 Exam Tip: Write "Let the required number be k" at the start. Show the substitution step clearly. Double-check your arithmetic with negative numbers.

Question 11. When the polynomial \( x^3 + 2x^2 - 5ax - 7 \) is divided by \( (x - 1) \), the remainder is A and when the polynomial \( x^3 + ax^2 - 12x + 16 \) is divided by \( (x + 2) \), the remainder is B. Find the value of 'a' if \( 2A + B = 0 \).
Answer:
Step 1: Find remainder A.
When \( x^3 + 2x^2 - 5ax - 7 \) is divided by \( (x - 1) \):
A = \( (1)^3 + 2(1)^2 - 5a(1) - 7 \)
A = \( 1 + 2 - 5a - 7 = -5a - 4 \) ...(i)

Step 2: Find remainder B.
When \( x^3 + ax^2 - 12x + 16 \) is divided by \( (x + 2) \):
B = \( (-2)^3 + a(-2)^2 - 12(-2) + 16 \)
B = \( -8 + 4a + 24 + 16 = 4a + 32 \) ...(ii)

Step 3: Use the condition \( 2A + B = 0 \).
Substituting from (i) and (ii):
\( 2(-5a - 4) + 4a + 32 = 0 \)
\( -10a - 8 + 4a + 32 = 0 \)
\( -6a + 24 = 0 \)
\( 6a = 24 \)
\( a = 4 \)

Answer: a = 4
In simple words: We find both remainders using the Remainder Theorem. Then we use the given condition to make an equation and solve for a.

📝 Teacher's Note: Show students to be very careful with signs when substituting negative values. Make them double-check each arithmetic step.

🎯 Exam Tip: Label your remainders clearly as A and B. Show the substitution step for both polynomials. Write the final equation \( 2A + B = 0 \) clearly.

Question 12. \( (3x + 5) \) is a factor of the polynomial \( (a - 1)x^3 + (a + 1)x^2 - (2a + 1)x - 15 \). Find the value of 'a', factorize the given polynomial completely.
Answer:
Step 1: Set up the condition for factor.
Let \( f(x) = (a - 1)x^3 + (a + 1)x^2 - (2a + 1)x - 15 \)
Since \( (3x + 5) \) is a factor, remainder = 0
Therefore, \( f\left(-\frac{5}{3}\right) = 0 \)

Step 2: Substitute \( x = -\frac{5}{3} \).
\( (a - 1)\left(-\frac{5}{3}\right)^3 + (a + 1)\left(-\frac{5}{3}\right)^2 - (2a + 1)\left(-\frac{5}{3}\right) - 15 = 0 \)

\( (a - 1)\left(-\frac{125}{27}\right) + (a + 1)\left(\frac{25}{9}\right) - (2a + 1)\left(-\frac{5}{3}\right) - 15 = 0 \)

Step 3: Simplify the equation.
\( -\frac{125(a - 1)}{27} + \frac{25(a + 1)}{9} + \frac{5(2a + 1)}{3} - 15 = 0 \)

Multiplying by 27:
\( -125(a - 1) + 75(a + 1) + 45(2a + 1) - 405 = 0 \)
\( -125a + 125 + 75a + 75 + 90a + 45 - 405 = 0 \)
\( 40a - 160 = 0 \)
\( 40a = 160 \)
\( a = 4 \)

Step 4: The polynomial becomes:
\( f(x) = 3x^3 + 5x^2 - 9x - 15 \)

Step 5: Complete factorization.
[Diagram: Long division showing the factorization steps]
\( 3x^3 + 5x^2 - 9x - 15 = (3x + 5)(x^2 - 3) = (3x + 5)(x + \sqrt{3})(x - \sqrt{3}) \)

Answer: a = 4
Complete factorization: \( (3x + 5)(x + \sqrt{3})(x - \sqrt{3}) \)
In simple words: If \( (3x + 5) \) is a factor, then putting \( x = -\frac{5}{3} \) makes the polynomial zero. We solve to find a.

📝 Teacher's Note: Show students how to find the zero of \( 3x + 5 = 0 \) gives \( x = -\frac{5}{3} \). Work through fraction arithmetic slowly.

🎯 Exam Tip: Always write the zero correctly: if \( (3x + 5) \) is a factor, then \( x = -\frac{5}{3} \). Show polynomial division for complete factorization.

Question 13. When divided by \( x - 3 \) the polynomials \( x^3 - px^2 + x + 6 \) and \( 2x^3 - x^2 - (p + 3)x - 6 \) leave the same remainder. Find the value of 'p'.
Answer:
Step 1: Find remainder for first polynomial.
If \( (x - 3) \) divides \( f(x) = x^3 - px^2 + x + 6 \):
Remainder = \( f(3) = 3^3 - p(3)^2 + 3 + 6 = 27 - 9p + 9 = 36 - 9p \)

Step 2: Find remainder for second polynomial.
If \( (x - 3) \) divides \( g(x) = 2x^3 - x^2 - (p + 3)x - 6 \):
Remainder = \( g(3) = 2(3)^3 - (3)^2 - (p + 3)(3) - 6 \)
= \( 54 - 9 - 3p - 9 - 6 = 30 - 3p \)

Step 3: Set remainders equal.
\( f(3) = g(3) \)
\( 36 - 9p = 30 - 3p \)
\( 36 - 30 = 9p - 3p \)
\( 6 = 6p \)
\( p = 1 \)

Answer: p = 1
In simple words: Both polynomials give the same remainder when divided by \( (x - 3) \). So we make their values at \( x = 3 \) equal.

📝 Teacher's Note: Remind students that "same remainder" means the polynomials have the same value when we substitute the root of the divisor.

🎯 Exam Tip: Write both remainder calculations clearly. Show the equation \( f(3) = g(3) \) and solve step by step.

Question 14. Use the Remainder Theorem to factorize the following expression: \( 2x^3 + x^2 - 13x + 6 \)
Answer:
Step 1: Find a factor by trial.
Let \( f(x) = 2x^3 + x^2 - 13x + 6 \)
Try factors of constant term 6: ±1, ±2, ±3, ±6

Step 2: Test \( x = 2 \).
\( f(2) = 2(2)^3 + (2)^2 - 13(2) + 6 = 16 + 4 - 26 + 6 = 0 \)
Hence \( (x - 2) \) is a factor of \( f(x) \).

Step 3: Divide by \( (x - 2) \).
[Diagram: Long division showing the steps]
\( 2x^3 + x^2 - 13x + 6 = (x - 2)(2x^2 + 5x - 3) \)

Step 4: Factor the quadratic.
\( 2x^2 + 5x - 3 = (2x - 1)(x + 3) \)

Complete factorization: \( 2x^3 + x^2 - 13x + 6 = (x - 2)(2x - 1)(x + 3) \)
In simple words: We try different values until we find one that makes the polynomial zero. Then we divide and factor the remaining part.

📝 Teacher's Note: Teach students to try factors of the constant term first. Show them the systematic way to test values.

🎯 Exam Tip: Always show the trial step where you test \( f(2) = 0 \). Then show polynomial division clearly. Check your final answer by expanding.

Question 15. Using remainder theorem, find the value of k if on dividing \( 2x^3 + 3x^2 - kx + 5 \) by \( x - 2 \), leaves a remainder 7.
Answer:
Step 1: Apply Remainder Theorem.
Let \( f(x) = 2x^3 + 3x^2 - kx + 5 \)
Using Remainder Theorem: \( f(2) = 7 \)

Step 2: Substitute \( x = 2 \).
\( 2(2)^3 + 3(2)^2 - k(2) + 5 = 7 \)
\( 16 + 12 - 2k + 5 = 7 \)
\( 33 - 2k = 7 \)
\( 2k = 26 \)
\( k = 13 \)

Answer: k = 13
In simple words: The remainder when dividing by \( (x - 2) \) is the value we get when we put \( x = 2 \) in the polynomial.

📝 Teacher's Note: Show students that remainder = 7 means \( f(2) = 7 \). This is the key step in using Remainder Theorem.

🎯 Exam Tip: Write "Using Remainder Theorem, \( f(2) = 7 \)" clearly. Show all arithmetic steps to avoid calculation errors.

Question 16. What must be subtracted from \( 16x^3 - 8x^2 + 4x + 7 \) so that the resulting expression has \( 2x + 1 \) as a factor?
Answer:
Step 1: Set up the problem.
Let \( f(x) = 16x^3 - 8x^2 + 4x + 7 \)
Let the number to be subtracted be k.
New polynomial = \( f(x) - k \)

Step 2: Apply the factor condition.
For \( (2x + 1) \) to be a factor, remainder must be 0 when divided by \( (2x + 1) \).
This means \( f\left(-\frac{1}{2}\right) - k = 0 \)

Step 3: Calculate \( f\left(-\frac{1}{2}\right) \).
\( f\left(-\frac{1}{2}\right) = 16\left(-\frac{1}{2}\right)^3 - 8\left(-\frac{1}{2}\right)^2 + 4\left(-\frac{1}{2}\right) + 7 \)
= \( 16\left(-\frac{1}{8}\right) - 8\left(\frac{1}{4}\right) + 4\left(-\frac{1}{2}\right) + 7 \)
= \( -2 - 2 - 2 + 7 = 1 \)

Step 4: Find k.
\( f\left(-\frac{1}{2}\right) - k = 0 \)
\( 1 - k = 0 \)
\( k = 1 \)

Answer: 1 must be subtracted.
In simple words: We need to subtract a number so that when we put \( x = -\frac{1}{2} \), the polynomial becomes zero.

📝 Teacher's Note: Show students that \( 2x + 1 = 0 \) gives \( x = -\frac{1}{2} \). Be very careful with fraction calculations and signs.

🎯 Exam Tip: Always find the zero of the factor first: \( 2x + 1 = 0 \) gives \( x = -\frac{1}{2} \). Show fraction arithmetic step by step.

Solution:
Here, \( f(x) = 16x^3 - 8x^2 + 4x + 7 \)
Let the number subtracted be k from the given polynomial f(x).

Given that \( 2x + 1 \) is a factor of \( f(x) \).

\( f\left(-\frac{1}{2}\right) = 0 \)

\( \Rightarrow 16\left(-\frac{1}{2}\right)^3 - 8\left(-\frac{1}{2}\right)^2 - 4\left(-\frac{1}{2}\right) - 7 - k = 0 \)

\( \Rightarrow 16 \times \left(-\frac{1}{8}\right) - 8 \times \frac{1}{4} - 2 - 7 - k = 0 \)

\( \Rightarrow -2 - 2 - 2 - 7 - k = 0 \)

\( \Rightarrow -6 - 7 - k = 0 \)

\( \Rightarrow k = 1 \)

Therefore 1 must be subtracted from \( 16x^3 - 8x^2 + 4x + 7 \) so that the resulting expression has \( 2x + 1 \) as a factor.

📝 Teacher's Note: When a polynomial has a factor \( (ax + b) \), then \( x = -\frac{b}{a} \) makes the polynomial equal to zero. This is called the factor theorem. Students often forget the negative sign.

🎯 Exam Tip: Always substitute \( x = -\frac{1}{2} \) when the factor is \( 2x + 1 \). Show each calculation step clearly. Write "k = 1" as your final answer.