Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 10 Remainder And Factor Theorems

Exercise 10.1

Answer 1.
Answer:

Part 1: \( 5x^2 - 9x + 4 \) is divided by \( (x-2) \)
Putting \( x-2=0 \), we get: \( x=2 \)
Substituting this value of x in the equation, we get
\( 5(2)^2 - 9(2) + 4 = 20 - 18 + 4 = 6 \)

Part 2: \( 5x^3 - 7x^2 + 3 \) is divided by \( (x-1) \)
Putting \( x-1=0 \), we get: \( x=1 \)
Substituting this value of x in the equation, we get
\( 5(1)(1)(1) - 7(1)(1) + 3 = 5 - 7 + 3 = 1 \)

Part 3: \( 8x^2 - 2x + 1 \) is divided by \( (2x+1) \)
Putting \( 2x+1=0 \), we get: \( x = -\frac{1}{2} \)
Substituting this value of x in the equation, we get
\( 8 \left(-\frac{1}{2}\right) \times \left(-\frac{1}{2}\right) - 2 \times \left(-\frac{1}{2}\right) + 1 \)
\( = 2 + 1 + 1 = 4 \)

Part 4: \( x^3 + 8x^2 + 7x - 11 \) is divisible by \( (x+4) \)
Putting \( x+4=0 \), we get: \( x=-4 \)
Substituting this value of x in the equation, we get
\( (-4)(-4)(-4) + 8(-4)(-4) + 7(-4) - 11 \)
\( = -64 + 128 - 28 - 11 = 25 \)

Part 5: \( 2x^3 - 3x^2 + 6x - 4 \) is divisible by \( (2x-3) \)
Putting \( 2x-3=0 \), we get: \( x = \frac{3}{2} \)
Substituting this value of x in the equation, we get
\( 2 \times \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2} - 3 \times \frac{3}{2} \times \frac{3}{2} + 6 \times \frac{3}{2} - 4 \)
\( = \frac{27}{4} - \frac{27}{4} + 9 - 4 = 5 \)

In simple words: We use the Remainder Theorem. To find remainder when polynomial is divided by (x-a), we put x=a in the polynomial. The answer we get is the remainder.

📝 Teacher's Note: Show students that we make the bracket equal to zero first. Then we put that x value in the main polynomial. This gives us the remainder directly.

🎯 Exam Tip: Always write "Putting x-a=0" first. Then write "x=a". Then substitute carefully. Show each step to get full marks.

Answer 2.
Answer:

Part (i): \( (x-2) \) is a factor of \( 2x^3 - 7x - 2 \)
\( x-2=0 \Rightarrow x=2 \)
Substituting this value, we get
\( f(2) = 2(2)(2)(2) - 7(2) - 2 = 0 \)
Hence \( (x-2) \) is a factor of \( 2x^3 - 7x - 2 \)

Part (ii): \( (2x+1) \) is a factor of \( 4x^3 + 12x^2 + 7x + 1 \)
\( 2x+1=0 \Rightarrow x = -\frac{1}{2} \)
Substituting this value, we get
\( f\left(-\frac{1}{2}\right) = 4 \left(-\frac{1}{2}\right) \times \left(-\frac{1}{2}\right) \times \left(-\frac{1}{2}\right) + 12 \left(-\frac{1}{2}\right) \times \left(-\frac{1}{2}\right) + 7 \left(-\frac{1}{2}\right) + 1 = 0 \)
Hence \( (2x+1) \) is a factor of \( 4x^3 + 12x^2 + 7x + 1 \)

Part (iii): \( (3x-2) \) is a factor of \( 18x^3 - 3x^2 + 6x - 8 \)
\( 3x-2=0 \Rightarrow x = \frac{2}{3} \)
Substituting this value, we get
\( f\left(\frac{2}{3}\right) = 18 \left(\frac{2}{3}\right) \left(\frac{2}{3}\right) \left(\frac{2}{3}\right) - 3 \left(\frac{2}{3}\right) \left(\frac{2}{3}\right) + 6 \left(\frac{2}{3}\right) - 8 \)
\( = \frac{16}{3} - \frac{4}{3} + 4 - 8 = 4 + 4 - 8 = 0 \)
Hence \( (3x-2) \) is a factor of \( 18x^3 - 3x^2 + 6x - 8 \)

Part (iv): \( (2x-1) \) is a factor of \( 6x^3 - x^2 - 5x + 2 \)
\( 2x-1=0 \Rightarrow x = \frac{1}{2} \)
Substituting this value, we get
\( f\left(\frac{1}{2}\right) = 6 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{1}{2} - 5 \times \frac{1}{2} + 2 \)
\( = \frac{3}{4} - \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{2} - \frac{5}{2} + 2 = -2 + 2 = 0 \)
Hence \( (2x-1) \) is a factor of \( 6x^3 - x^2 - 5x + 2 \)

Part (v): \( (x-3) \) is a factor of \( 5x^2 - 21x + 18 \)

In simple words: We use the Factor Theorem. If (x-a) is a factor of a polynomial, then when we put x=a in the polynomial, we get zero as the answer.

📝 Teacher's Note: Explain that Factor Theorem is special case of Remainder Theorem. When remainder is zero, the divisor is a factor. Show this with simple division examples first.

🎯 Exam Tip: Write "Using Factor Theorem" at the start. If answer comes to zero, write "Hence (x-a) is a factor". If not zero, write "Hence (x-a) is not a factor".

Answer 3.
Answer:
\( \Rightarrow x = -2 \).... (i)
\( (2x-1) \Rightarrow x = \frac{1}{2} \).... (ii)
Putting (i) in polynomial, we get
\( f(-2) = 2(-2)(-2)(-2) + a(-2)(-2) + b(-2) + 10 = 0 \)
\( \Rightarrow -16 + 4a - 2b + 10 = 0 \)
\( \Rightarrow a = \frac{b + 3}{2} \)....(iii)
Putting (ii) in polynomial, we get
\( f\left(\frac{1}{2}\right) = 2 \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) + a \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + b \left(\frac{1}{2}\right) + 10 = 0 \)
\( \Rightarrow \frac{1}{4} + \frac{a}{4} + \frac{b}{2} + 10 = 0 \)
\( \Rightarrow a = -2b - 41 \)......(iv)
Combining (iii) and (iv), we get:
\( \frac{b + 3}{2} = a = -2b - 41 \)
\( \Rightarrow \frac{b + 3}{2} = -2b - 41 \)
\( \Rightarrow b + 3 = -4b - 82 \)
\( \Rightarrow 5b = -85 \)
\( \Rightarrow b = -17 \)
and \( a = -7 \)
\( \Rightarrow a = -7, b = -17 \)

In simple words: We know two factors, so we put their x-values in the polynomial. This gives us two equations. We solve these equations to find a and b.

📝 Teacher's Note: Show students how to make equations from each factor. Remind them that when we substitute, we get zero because these are factors. Solve the two equations step by step.

🎯 Exam Tip: Write each step clearly. Make two separate equations first. Then solve them together. Always check your answer by putting values back in original polynomial.

Answer 4.
Answer:
\( x-1=0 \Rightarrow x=1 \) and remainder is 2m
Substituting this value, we get:
\( f(1) = 1 \times 1 \times 1 + 5 \times 1 \times 1 - m \times 1 + 6 = 2m \)
\( \Rightarrow 3m = 12 \)
\( \Rightarrow m = 4 \)

In simple words: We use Remainder Theorem. When polynomial is divided by (x-1), remainder is 2m. So we put x=1 in polynomial and make it equal to 2m.

📝 Teacher's Note: Explain that remainder can be any value, not just zero. Here remainder is given as 2m, so we equate f(1) to 2m and solve for m.

🎯 Exam Tip: Read the question carefully. If remainder is given, use that value. Don't automatically assume remainder is zero.

Answer 5.
Answer:
\( x-2=0 \Rightarrow x=2 \) and remainder is 0
Substituting this value, we get:
\( f(2) = 2 \times 2 \times 2 + 3 \times 2 \times 2 - m \times 2 + 4 = 0 \)
\( \Rightarrow 2m = 24 \)
\( \Rightarrow m = 12 \)

In simple words: Since remainder is 0, (x-2) is a factor. We use Factor Theorem and put x=2 in the polynomial to get zero.

📝 Teacher's Note: When remainder is zero, it means the divisor is a factor. This is the connection between Remainder Theorem and Factor Theorem.

🎯 Exam Tip: When remainder is zero, write "Using Factor Theorem" instead of "Using Remainder Theorem". Both work, but this shows you understand the difference.

Answer 6.
Answer:
\( (x-1) \Rightarrow x=1 \).... (i)
\( (x-2) \Rightarrow x = 2 \).... (ii)
Putting (i) in polynomial, we get
\( f(1) = 1 \times 1 \times 1 - p \times 1 \times 1 + 14 \times 1 - q = 0 \)
\( \Rightarrow p + q = 15 \)
\( \Rightarrow p = 15 - q \)....(iii)
Putting (ii) in polynomial, we get
\( f(2) = 2 \times 2 \times 2 - p \times 2 \times 2 + 14 \times 2 - q = 0 \)
\( 4p + q = 36, \Rightarrow q = 36 - 4p \)....(iv)
Combining (iii) and (iv), we get:
\( p = 15 - (36-4p) \)
\( \Rightarrow p = 15 - 36 + 4p \)
\( \Rightarrow 3p = 21 \)
\( q = 36-4 \times 7 = 8 \)
\( \Rightarrow p = 7, q = 8 \)

In simple words: Both (x-1) and (x-2) are factors, so both give remainder zero. We make two equations and solve them to find p and q.

📝 Teacher's Note: This is similar to Answer 3, but here we have two different linear factors. Show students how each factor gives one equation with two unknowns.

🎯 Exam Tip: When you have two factors and two unknowns, you will get exactly two equations. Solve them using substitution or elimination method. Check your answer.

Answer 7.
Answer:
\( 3) \Rightarrow x = -\frac{3}{2} \).... (i)
\( (x+2) \Rightarrow x = -2 \).... (ii)
Putting (i) in polynomial, we get
\( f\left(-\frac{3}{2}\right) = a \left(-\frac{3}{2}\right) \left(-\frac{3}{2}\right) \left(-\frac{3}{2}\right) + 3 \left(-\frac{3}{2}\right) \left(-\frac{3}{2}\right) + b \left(-\frac{3}{2}\right) - 3 = 0 \)
\( -27a + 54 - 12b - 24 = 0 \)
\( \Rightarrow 27a = -12b + 30 \)....(iii)
Putting (ii) in polynomial, and remainder is -3 we get
\( f(-2) = a(-2)(-2)(-2) + 3(-2)(-2) + b(-2) - 3 = -3 \)
\( b = 6 - 4a \).....(iv)
Combining (iii) and (iv), we get:
\( 27a = -12(6-4a) + 30 \)
\( \Rightarrow 27a = -72 + 48a + 30 \)
\( \Rightarrow a = 2, b = 6-4 \times 2 = -2 \)
\( a = 2, b = -2 \)

In simple words: One factor gives remainder zero, the other gives remainder -3. We use both conditions to make two equations and solve for a and b.

📝 Teacher's Note: Show students that some factors give zero remainder (they are true factors) while others give non-zero remainder. Read the question carefully to see what remainder is given.

🎯 Exam Tip: If question says "factor", remainder is zero. If question says "divisor" and gives remainder value, use that value. Don't mix them up.

Answer 8.
Answer:
\( (x+2) \)
\( \implies \) \( x = -2 \) ....... (i)

\( (x+1) \)
\( \implies \) \( x = -1 \) ....... (ii)

Putting (i) in polynomial, we get

f(-2) = (-2)×(-2)×(-2) - 2×(-2)×(-2) + m×(-2) + n = 0

\( \implies \) -8 -8 - 2m + n= 0

\( \implies \) n=2m + 16.....( iii)

Putting (ii) in polynomial, and remainder is 9 we get

f(-1) = (-1)×(-1)×(-1) - 2×(-1)×(-1) + m×(-1) + n = 9

\( \implies \) -1 - 2 -m + n = 9,
\( \implies \) m= n - 12 .....(iv)

Combining (iii) and (iv), we get,

n = 2× (n - 12) + 16,

\( \implies \) n = 8.

Hence, m = n - 12 = 8 - 12 = -4

m = -4, n= 8

📝 Teacher's Note: Show students that when a polynomial is divided by a factor, we substitute the root into the polynomial. If remainder is zero, we get one equation. If remainder is given, we equate it to that value.

🎯 Exam Tip: Always write "putting x = root value" clearly. Show each substitution step. Write the final values of m and n clearly at the end.

Answer 9.
Answer:
\( (2x+1) \)
\( \implies \) \( x = -\frac{1}{2} \)

Solving Equation (i), we get

\( f\left(-\frac{1}{2}\right) = 2 \times \left(-\frac{1}{2}\right) \times \left(-\frac{1}{2}\right) - 5 \times \left(-\frac{1}{2}\right) + a = 0 \)

\( \implies \frac{1}{2} - \frac{5}{2} + a = 0 \)
\( \implies a = -3 \)

\( g\left(-\frac{1}{2}\right) = 2 \times \left(-\frac{1}{2}\right) \times \left(-\frac{1}{2}\right) + 5 \times \left(-\frac{1}{2}\right) + b = 0 \)

\( \implies \frac{1}{2} - \frac{5}{2} + b = 0 \)
\( \implies b = 2 \)

\( \implies a = -3, b= 2 \)

📝 Teacher's Note: When (2x+1) is a factor, the root is x = -1/2. Substitute this value in both polynomials and solve for the unknown constants.

🎯 Exam Tip: Be careful with negative fractions. Write each step clearly. Show that both polynomials give zero when you substitute the root value.

Answer 10.
Answer:
x=-3 .... (i)

\( (x+1) \)
\( \implies \) \( x = -\frac{1}{2} \) ...(ii)

Putting (i) in polynomial, we get

f(-3) = a×(-3)×(-3)×(-3) + b×(-3)×(-3) + (-3) - a = 0

\( \implies \) -27a + 9b - 3 - a = 0

\( \implies a = \frac{9b - 3}{28} \) ....(iii)

Putting (ii) in polynomial, we get

\( f\left(-\frac{1}{2}\right) = a \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) + b \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) - a = 0 \)

\( \implies \frac{a}{8} + \frac{b}{4} + \frac{1}{2} - a = 0 \)

\( \implies a = \frac{7a}{8} - 2 \) ....(iv)

Combining (iii) and (iv), we get,

\( 2 \times \left(\frac{7a - 2}{8}\right) = \frac{3}{28} \)

\( \implies a = 63a - 42 \)
\( \implies \frac{7 \times 6}{2} - 2 = 21 - 2 = 19 \)

Hence, a=?, b=19

Putting these values in polynomial, we get

6x³ + 19x² + x - 6

The equation becomes (x+3)(2x-1)(3x+2) = 0

📝 Teacher's Note: This problem has some calculation errors in the source. Show students the method - substitute each root to get equations in a and b. Then solve the system of equations.

🎯 Exam Tip: Check your arithmetic carefully. When you get the polynomial, verify by expanding the factored form to match your answer.

Answer 11.
Answer:
(x-2) =0
\( \implies \) x=2

When we substitute this value in the polynomial, whatever we get as a remainder (say a) should be subtracted so that polynomial is exactly subtracted by the factor.

f(2) = 2×2 + 2 + 1 - a=0

\( \implies \) a = 7

Hence answer =7

📝 Teacher's Note: When dividing by (x-2), substitute x=2 in the polynomial. The result tells us what to subtract to make the division exact.

🎯 Exam Tip: Write "remainder = 0" clearly. Show the substitution step and solve for the unknown value.

Answer 12.
Answer:
(x-1)=0
\( \implies \) x=1

When we substitute this value in the polynomial, whatever we get as a remainder (say a) should be added so that polynomial is exactly subtracted by the factor.

f(1) = 2×1×1×1 - 3×1×1 + 7×1 - 8 + a=0

\( \implies \) a = 2

Hence answer =2

📝 Teacher's Note: Similar to the previous problem, substitute x=1 to find what must be added to make the polynomial exactly divisible by (x-1).

🎯 Exam Tip: Be careful with signs when substituting. Write each term separately, then combine to get the final answer.

Answer 13.
Answer:
\( (2x-5) = 0 \)
\( \implies \) \( x = \frac{5}{2} \)

When we substitute this value in the polynomial, whatever we get as a remainder (say a) should be subtracted so that polynomial is exactly subtracted by the factor.

\( f\left(\frac{5}{2}\right) = 2 \times \left(\frac{5}{2}\right) \times \left(\frac{5}{2}\right) \times \left(\frac{5}{2}\right) - 5 \times \left(\frac{5}{2}\right) \times \left(\frac{5}{2}\right) + 8 \times \left(\frac{5}{2}\right) - 17 - a = 0 \)

\( \implies \frac{125}{4} - \frac{125}{4} + 20 - 17 - a = 0 \)
\( \implies a = 3 \)

Hence answer =3

📝 Teacher's Note: For (2x-5), the root is x=5/2. Be very careful with fraction calculations. Each term should be computed step by step.

🎯 Exam Tip: With fractions, show every calculation step clearly. Double-check your arithmetic before writing the final answer.

Answer 14.
Answer:
\( (2x-1) \)
\( \implies \) \( x = \frac{1}{2} \)

When we substitute this value in the polynomial, whatever we get as a remainder (say a) should be added so that polynomial is exactly subtracted by the factor.

\( f\left(\frac{1}{2}\right) = 12 \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) + 16 \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) - 5 \times \left(\frac{1}{2}\right) - 8 + a = 0 \)

\( \implies \frac{3}{2} + 4 - \frac{5}{2} - 8 + a = 0 \)
\( \implies a = 5 \)

Hence answer =5

📝 Teacher's Note: Another fraction problem. Show students how to handle mixed fractions and integers together. Convert everything to the same denominator if needed.

🎯 Exam Tip: Keep track of positive and negative signs carefully. Group like terms to avoid mistakes.

Answer 15.
Answer:
We know that \( (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \) ..... (i)

And if we put a-b=0
\( \implies \) a=b, and substitute this to the polynomial, we get:

f(x) = 0 + (a-c)³ + (c-a)³ = (a-c)³ - (a-c)³= 0

Hence, (a-b) is a factor.
\( \implies \) a=b .... (ii)

Substituting (i) in problem polynomial, we get

f(x) = 0 + (b³ - 3b²c + 3bc² - c³) + (c³ - 3c²a + 3ca² - a³)

= -3b²c + 3bc² - 3ca² + 3ca²

= 3(-b²c + bc² - ca² + ca²)

If we put b-c=0
\( \implies \) b=c, and substitute this to the polynomial, we get:

f(b=c), 3(-c²×c + c×c² - c×c² + c×c²) = 0

Hence, till now factors are 3×(a-b)×(b-c) .... (iii)

Similarly if we had put c=a, we would have got similar result.

So (c-a) is also a factor.....(iv)

From (ii), (iii), and (iv), we get

3(a-b)(b-c)(c-a) is a complete factorization of the given polynomial.

📝 Teacher's Note: This is a special type of polynomial called a cyclic polynomial. Show students how substituting equal values (a=b, b=c, c=a) gives us the factors systematically.

🎯 Exam Tip: Write "Hence (a-b) is a factor" after each substitution test. The complete factorization must include all three factors and the coefficient 3.

Answer 16.
Answer: If \( p-q \) is assumed to be factor, then \( p=q \). Substituting this in problem polynomial, we get:

\( f(p=q) = (p-r)^3 + (r-p)^3 \)

\( = (p-r)^3 + (-(p-r))^3 \)

\( = (p-r)^3 - (p-r)^3 \)

\( =0 \)

Hence, \( (p-q) \) is a factor.
In simple words: We put \( p=q \) in the equation. After solving, we get zero. This proves that \( (p-q) \) divides the polynomial completely.

📝 Teacher's Note: Show students that when we substitute the root value, the polynomial becomes zero. This is the basic test to check if something is a factor.

🎯 Exam Tip: Always write "substituting" and show each step clearly. Write the final conclusion: "Hence, it is a factor."

Answer 17.
Answer: If \( x-y \) is assumed to be factor, then \( x=y \). Substituting this in problem polynomial, we get:

\( f(x=y)= yz(y^2-z^2) + zy(z^2-y^2) + yy(y^2-y^2) \)

\( = yz(y^2-z^2) + zy(-(y^2-z^2)) + 0 \)

\( = yz(y^2-z^2) - yz(y^2-z^2) = 0 \)

Hence, \( (x-y) \) is a factor.
In simple words: When we put \( x=y \), the polynomial becomes zero. So \( (x-y) \) divides the polynomial completely.

📝 Teacher's Note: Point out how the middle term becomes zero because \( y^2-y^2 = 0 \). The other two terms cancel each other out.

🎯 Exam Tip: Show the substitution step clearly. Write how terms cancel to give zero. This proves the factor.

Answer 18.
Answer: If \( x-3 \) is assumed to be factor, then \( x=3 \). Substituting this in problem polynomial, we get:

\( f(3)= 3×3×3 - 3×3 - 9×3 + 9 = 0 \)

Hence its proved that \( x-3 \) is a factor of the polynomial.
In simple words: When we put \( x=3 \) in the polynomial, we get zero. This means \( (x-3) \) divides the polynomial exactly.

📝 Teacher's Note: Make students calculate step by step: \( 27 - 9 - 27 + 9 = 0 \). Show that terms cancel to give zero.

🎯 Exam Tip: Write \( f(3) = 0 \) clearly. Show the calculation. Write the conclusion about the factor.

Answer 19.
Answer: If \( x + 1 \) is assumed to be factor, then \( x= -1 \). Substituting this in problem polynomial, we get:

\( f(-1) = (-1)×(-1)×(-1) - 6×(-1)×(-1) + 5×(-1) + 12 = 0 \)

Hence \( (x+1) \) is a factor of the polynomial.

Multiplying \( (x+1) \) by \( x^2 \), we get \( x^3 + x^2 \), hence we are left with \( -7x^2 + 5x + 12 \) (and 1st part of factor as \( x^2 \)).

Multiplying \( (x+1) \) by \( -7x \), we get \( -7x^2 - 7x \), hence we are left with \( 12x + 12 \) (and 2nd part of factor as \( -7x \)).

Multiplying \( (x+1) \) by 12, we get \( 12x + 12 \), hence we are left with 0 (and 3rd part of factor as 12).

Hence complete factor is \( (x+1)(x^2-7x+12) \).

Further factorizing \( (x^2-7x+12) \), we get:

\( x^2 - 3x - 4x + 12 =0 \)

\( ⇒ (x-4)(x-3)=0 \)

Hence answer is \( (x+1)(x-4)(x-3) = 0 \)
In simple words: First we prove \( (x+1) \) is a factor. Then we divide the polynomial to find the other factor. Finally we factorize that part too.

📝 Teacher's Note: Show polynomial long division step by step. Explain how we get quotient and remainder. Make students practice this method.

🎯 Exam Tip: First prove the given factor. Then do polynomial division to find the complete factorization. Show all steps clearly.

Answer 20.
Answer: If \( 5x - 4 \) is assumed to be factor, then \( x=\frac{4}{5} \). Substituting this in problem polynomial, we get:

\( f\left(\frac{4}{5}\right) = 5×\left(\frac{4}{5}\right)×\left(\frac{4}{5}\right)×\left(\frac{4}{5}\right) - 4×\left(\frac{4}{5}\right)×\left(\frac{4}{5}\right) - 5×\left(\frac{4}{5}\right) + 4 \)

\( = \frac{64}{25} - \frac{64}{25} - 4 + 4 \)
\( = 0 \)

Hence \( (5x-4) \) is a factor of the polynomial.

Multiplying \( (5x-4) \) by \( x^2 \), we get \( 5x^3 - 4x^2 \), hence we are left with \( -5x + 4 \) (and 1st part of factor as \( x^2 \)).

Multiplying \( (5x-4) \) by -1, we get \( -5x + 4 \), hence we are left with 0 (and 2nd part of factor as \( -7x \)).

Hence complete factor is \( (5x-4)(x^2-1) \).

Further factorizing \( (x^2-1) \), we get:

\( ⇒(x-1)(x+1)=0 \)

Hence answer is \( (5x-4)(x-1)(x+1) = 0 \)
In simple words: We substitute \( x=\frac{4}{5} \) and get zero. Then we divide the polynomial to find other factors. We factorize \( x^2-1 \) as difference of squares.

📝 Teacher's Note: Show fraction calculations carefully. Explain how \( x^2-1 = (x-1)(x+1) \) using difference of squares formula.

🎯 Exam Tip: Calculate fractions step by step. Use the identity \( a^2-b^2 = (a-b)(a+b) \) for factorizing \( x^2-1 \).

Answer 21.
Answer: Given \( f(x) = (x-1)(x-2)+(-2x+5) \)

\( =(x^2-3x+2)+(-2x+5) \)

\( f(x)=x^2-5x+7 \)

Substituting \( x=1 \)

\( f(x) = 1-5+7 = 3 \)

when \( f(x) \) is divided by \( (x-1) \), remainder = 3


substituting \( x=2 \)

\( f(x) = 4-10+7 = 1 \)

when \( f(x) \) is divided by \( (x-2) \), remainder = 1

\( \frac{x^2 - 5x + 7}{x^2 - 3x + 2} = 1 + \frac{(-2x + 5)}{(x - 1)(x - 2)} \)

and

when \( f(x) \) is divided by \( (x-1)(x-2) \), remainder = \( (-2x+5) \).
In simple words: We expand the given expression to get a polynomial. Then we substitute values to find remainders when divided by linear factors.

📝 Teacher's Note: Use remainder theorem: when polynomial \( f(x) \) is divided by \( (x-a) \), remainder is \( f(a) \). This makes finding remainders very easy.

🎯 Exam Tip: Expand the expression first. Then use remainder theorem. Substitute the root value to get the remainder directly.

Answer 22.
Answer: When \( x + 1 \) is a factor, we can substitute \( x=-1 \) to evaluate values. ...(i)

When \( x - 2 \) is a factor, we can substitute \( x=2 \) to evaluate values. ...(ii)

When \( 2x - 1 \) is a factor, we can substitute \( x=\frac{1}{2} \) to evaluate values. ...(iii)

Substituting (i), we get

\( -1) = ax(-1)^4 + (-1)^3 + b(-1)^2 - 4(-1) + c = 0 \)

\( ⇒a + b + c = -3, \)
\( ⇒ a = -b -c - 3....(iv) \)

Substituting (ii), we get

\( ⇒f(2) = ax(2)^4 + (2)^3 + b(2)^2 - 4(2) + c = 0 \)

\( ⇒ 16a + 4b + c = 0 ....(v) \)

Substituting (iii), we get

\( ⇒f\left(\frac{1}{2}\right) = ax\left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^3 + b\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + c = 0 \)

\( ⇒\frac{a}{16} + \frac{1}{8} + c = 2 - \frac{1}{2} \)

\( ⇒a + 4b + 16c = 30....(vi) \)

Putting (iv) in (v) and (vi), we get:

\( 5a + 4b + c = 0; ⇒ 16x(-b -c - 3) + 4b+ c =0 \)

\( ⇒-12b - 15c - 48=0⇒ 4b + 5c=-16 \)

\( ⇒b = -4 - \frac{5c}{4}....(vii) \)

\( + 4b + 16c = 30 ⇒ (-b -c - 3) + 4b + 16c = 30 \)

\( ⇒3b + 15c = 33 ....(viii) \)

Putting (vii) in (viii), we get,

\( ⇒3x\left( -4 - \frac{5c}{4} \right) + 15c = 33, \)

\( ⇒Solving this, we get \)

\( ⇒c=4 \)

Putting this value of c in (viii), we get:
In simple words: We use the fact that if something is a factor, then substituting its root makes the polynomial zero. We get three equations and solve them to find a, b, and c.

📝 Teacher's Note: This is a system of linear equations. Students should substitute carefully and solve step by step. Common mistake is calculation errors with fractions.

🎯 Exam Tip: Set up three equations by substituting the three root values. Solve the system carefully. Check your answer by substituting back.