Maharashtra Board Class 9 Maths Chapter 3 Set 3.5 Algebra Standard Part 1 Polynomials Solutions

Question 1. Find the value of the polynomial \( 2x - 2x^3 + 7 \) using given values for x.
i. \( x = 3 \)
ii. \( x = -1 \)
iii. \( x = 0 \)
Answer:
Let \( p(x) = 2x - 2x^3 + 7 \)

i. For \( x = 3 \):
Substitute \( x = 3 \) in the polynomial:
\( p(3) = 2(3) - 2(3)^3 + 7 \)

\( \implies p(3) = 6 - 2(27) + 7 \)

\( \implies p(3) = 6 - 54 + 7 \)

\( \implies p(3) = -48 + 7 \)

\( \implies p(3) = -41 \)
The value of the polynomial for \( x = 3 \) is \( -41 \).

ii. For \( x = -1 \):
Substitute \( x = -1 \) in the polynomial:
\( p(-1) = 2(-1) - 2(-1)^3 + 7 \)

\( \implies p(-1) = -2 - 2(-1) + 7 \)

\( \implies p(-1) = -2 + 2 + 7 \)

\( \implies p(-1) = 7 \)
The value of the polynomial for \( x = -1 \) is \( 7 \).

iii. For \( x = 0 \):
Substitute \( x = 0 \) in the polynomial:
\( p(0) = 2(0) - 2(0)^3 + 7 \)

\( \implies p(0) = 0 - 0 + 7 \)

\( \implies p(0) = 7 \)
The value of the polynomial for \( x = 0 \) is \( 7 \).
In simple words: To find the value of a polynomial, we just replace the variable x with the given number and calculate the final answer using basic arithmetic.

🎯 Exam Tip: Always pay close attention to negative signs when substituting values, especially when raising negative numbers to odd powers like \( (-1)^3 = -1 \).

Question 1. For the polynomial \( p(x) = 2x - 2x^3 + 7 \), find the value of the polynomial when \( x = 3 \), \( x = -1 \) and \( x = 0 \).
Answer:
(i) For \( x = 3 \):
\( p(x) = 2x - 2x^3 + 7 \)
Put \( x = 3 \) in the given polynomial.
\( \therefore p(3) = 2(3) - 2(3)^3 + 7 \)
\( = 6 - 2 \times 27 + 7 \)
\( = 6 - 54 + 7 \)
\( \therefore p(3) = -41 \)

(ii) For \( x = -1 \):
\( p(x) = 2x - 2x^3 + 7 \)
Put \( x = -1 \) in the given polynomial.
\( \therefore p(-1) = 2(-1) - 2(-1)^3 + 7 \)
\( = -2 - 2(-1) + 7 \)
\( = -2 + 2 + 7 \)
\( \dots p(-1) = 7 \)

(iii) For \( x = 0 \):
\( p(x) = 2x - 2x^3 + 7 \)
Put \( x = 0 \) in the given polynomial.
\( \therefore p(0) = 2(0) - 2(0)^3 + 7 \)
\( = 0 - 0 + 7 \)
\( \therefore p(0) = 7 \)
This process of substitution helps us determine the specific value of the algebraic expression at these points.
In simple words: To find the value of a polynomial for any number, we just replace the variable with that number and calculate the final answer.

🎯 Exam Tip: Be very careful with negative signs when raising a negative number to an odd power, as \( (-1)^3 \) remains \( -1 \).

Question 2. For each of the following polynomials, find \( p(1) \), \( p(0) \) and \( p(-2) \).
(i) \( p(x) = x^3 \)
(ii) \( p(y) = y^2 - 2y + 5 \)
(iii) \( p(x) = x^4 - 2x^2 + x \)
Answer:
(i) \( p(x) = x^3 \)
For \( p(1) \):
\( p(1) = 1^3 = 1 \)
For \( p(0) \):
\( p(0) = 0^3 = 0 \)
For \( p(-2) \):
\( p(-2) = (-2)^3 = -8 \)

(ii) \( p(y) = y^2 - 2y + 5 \)
For \( p(1) \):
\( p(1) = 1^2 - 2(1) + 5 \)
\( = 1 - 2 + 5 \)
\( \therefore p(1) = 4 \)
For \( p(0) \):
\( p(0) = 0^2 - 2(0) + 5 \)
\( = 0 - 0 + 5 \)
\( \therefore p(0) = 5 \)
For \( p(-2) \):
\( p(-2) = (-2)^2 - 2(-2) + 5 \)
\( = 4 + 4 + 5 \)
\( \therefore p(-2) = 13 \)

(iii) \( p(x) = x^4 - 2x^2 + x \)
For \( p(1) \):
\( p(1) = 1^4 - 2(1)^2 + 1 \)
\( = 1 - 2 + 1 \)
\( \therefore p(1) = 0 \)
For \( p(0) \):
\( p(0) = 0^4 - 2(0)^2 + 0 \)
\( = 0 - 0 + 0 \)
\( \dots p(0) = 0 \)
For \( p(-2) \):
\( p(-2) = (-2)^4 - 2(-2)^2 + (-2) \)
\( = 16 - 2(4) - 2 \)
\( = 16 - 8 - 2 \)
\( \therefore p(-2) = 6 \)
Evaluating these expressions systematically ensures we find the correct numerical value for each given input.
In simple words: We substitute the values 1, 0, and -2 into each polynomial one by one to find their respective values.

🎯 Exam Tip: Show each substitution step clearly and double-check your arithmetic, especially when dealing with exponents of negative numbers.

Question 2. Find the values of the polynomials as specified.
Answer: For the polynomial \( p(y) = y^2 - 2y + 5 \):
\( \implies p(0) = 0^2 - 2(0) + 5 \)
\( = 0 - 0 + 5 \)
\( \implies p(0) = 5 \)

For the polynomial \( p(y) = y^2 - 2y + 5 \):
\( \implies p(-2) = (-2)^2 - 2(-2) + 5 \)
\( = 4 + 4 + 5 \)
\( \implies p(-2) = 13 \)

(iii) For the polynomial \( p(x) = x^4 - 2x^2 - x \):
\( \implies p(1) = (1)^4 - 2(1)^2 - 1 \)
\( = 1 - 2 - 1 \)
\( \implies p(1) = -2 \)

\( \implies p(x) = x^4 - 2x^2 - x \)
\( \implies p(0) = (0)^4 - 2(0)^2 - 0 \)
\( = 0 - 0 - 0 \)
\( \implies p(0) = 0 \)

\( \implies p(x) = x^4 - 2x^2 - x \)
\( \implies p(-2) = (-2)^4 - 2(-2)^2 - (-2) \)
\( = 16 - 2(4) + 2 \)
\( = 16 - 8 + 2 \)
\( \implies p(-2) = 10 \)
In simple words: To find the value of a polynomial for any given number, we just replace the variable with that number and calculate the final answer.

🎯 Exam Tip: Be very careful with negative signs when squaring or cubing numbers, as a negative number raised to an even power always becomes positive.

Question 3. If the value of the polynomial \( m^3 + 2m + a \) is 12 for \( m = 2 \), then find the value of \( a \).
Answer: Let \( p(m) = m^3 + 2m + a \)
Given that for \( m = 2 \), the value of the polynomial is 12.
\( \implies p(2) = 12 \)
\( \implies (2)^3 + 2(2) + a = 12 \)
\( \implies 8 + 4 + a = 12 \)
\( \implies 12 + a = 12 \)
\( \implies a = 12 - 12 \)
\( \implies a = 0 \)
The value of \( a \) is determined to be zero.
In simple words: We substitute \( m = 2 \) into the expression, set the whole equation equal to 12, and solve for \( a \) by simplifying the numbers.

🎯 Exam Tip: Always write the reason like \( p(2) = 12 \) in brackets next to your step to show the examiner exactly how you substituted the value.

Question 4. For the polynomial \( mx^2 - 2x + 3 \) if \( p(-1) = 7 \), then find \( m \).
Answer: Let \( p(x) = mx^2 - 2x + 3 \)
Given that \( p(-1) = 7 \)
\( \implies m(-1)^2 - 2(-1) + 3 = 7 \)
\( \implies m(1) + 2 + 3 = 7 \)
\( \implies m + 5 = 7 \)
\( \implies m = 7 - 5 \)
\( \implies m = 2 \)
This calculation clearly shows that the coefficient \( m \) must be equal to 2.
In simple words: Replace \( x \) with \( -1 \) in the polynomial, set the entire expression equal to 7, and then solve for \( m \).

🎯 Exam Tip: Remember that \( (-1)^2 = 1 \) and \( -2 \times (-1) = +2 \). Double-check your signs to avoid simple calculation errors.

Question 5. Divide the first polynomial by the second polynomial and find the remainder using remainder theorem.
(i) \( (x^2 - 7x + 9); (x + 1) \)
(ii) \( (2x^3 - 2x^2 + ax - a); (x - a) \)
(iii) \( (54m^3 + 18m^2 - 27m + 5); (m - 3) \)
Answer:
(i)
Let \( p(x) = x^2 - 7x + 9 \)
Divisor = \( x + 1 \)

\( \implies \) take \( x = -1 \)
By remainder theorem,
Remainder = \( p(-1) \)
\( p(x) = x^2 - 7x + 9 \)

\( \implies p(-1) = (-1)^2 - 7(-1) + 9 \)
\( = 1 + 7 + 9 \)

\( \implies \text{Remainder} = 17 \)
This method is highly efficient as it avoids the lengthy process of polynomial long division.

(ii)
Let \( p(x) = 2x^3 - 2x^2 + ax - a \)
Divisor = \( x - a \)

\( \implies \) take \( x = a \)
By remainder theorem,
Remainder = \( p(a) \)
\( p(x) = 2x^3 - 2x^2 + ax - a \)

\( \implies p(a) = 2a^3 - 2a^2 + a(a) - a \)
\( = 2a^3 - 2a^2 + a^2 - a \)

\( \implies \text{Remainder} = 2a^3 - a^2 - a \)

(iii)
Let \( p(m) = 54m^3 + 18m^2 - 27m + 5 \)
Divisor = \( m - 3 \)

\( \implies \) take \( m = 3 \)
By remainder theorem,
Remainder = \( p(3) \)
\( p(m) = 54m^3 + 18m^2 - 27m + 5 \)

\( \implies p(3) = 54(3)^3 + 18(3)^2 - 27(3) + 5 \)
\( = 54(27) + 18(9) - 81 + 5 \)
\( = 1458 + 162 - 81 + 5 \)

\( \implies \text{Remainder} = 1544 \)
In simple words: To find the remainder when dividing a polynomial by a term like \( x - c \), we just plug the value \( c \) directly into the polynomial instead of doing long division.

🎯 Exam Tip: Always remember to change the sign of the constant in the divisor when substituting (e.g., if divisor is \( x + 1 \), substitute \( x = -1 \)) to avoid calculation errors.

Question 6. If the polynomial \( y^3 - 5y^2 + 7y + m \) is divided by \( y + 2 \) and the remainder is 50, then find the value of m.
Answer:
Let \( p(y) = y^3 - 5y^2 + 7y + m \)
Divisor = \( y + 2 \)
\( \therefore \) take \( y = -2 \)
\( \dots \) By remainder theorem,
Remainder = \( p(-2) = 50 \)
\( p(y) = y^3 - 5y^2 + 7y + m \)
\( \therefore p(-2) = (-2)^3 - 5(-2)^2 + 7(-2) + m \)
\( \therefore 50 = -8 - 5(4) - 14 + m \)
\( \therefore 50 = -8 - 20 - 14 + m \)
\( \therefore 50 = -42 + m \)
\( \therefore m = 50 + 42 \)
\( \therefore m = 92 \)
In simple words: We substitute \( y = -2 \) into the polynomial because the divisor is \( y + 2 \). Since the remainder is 50, we set the equation equal to 50 and solve to find that \( m \) is 92.

🎯 Exam Tip: Always remember to change the sign of the constant in the divisor when substituting (e.g., for \( y + 2 \), substitute \( y = -2 \)). Show each step of simplification clearly to avoid calculation errors.

Question 7. Use factor theorem to determine whether \( x + 3 \) is a factor of \( x^2 + 2x - 3 \) or not.
Answer:
Let \( p(x) = x^2 + 2x - 3 \)
Divisor = \( x + 3 \)
\( \dots \) take \( x = -3 \)
\( \therefore \) Remainder = \( p(-3) \)
\( p(x) = x^2 + 2x - 3 \)
\( \therefore p(-3) = (-3)^2 + 2(-3) - 3 \)
\( = 9 - 6 - 3 \)
\( \therefore p(-3) = 0 \)
\( \therefore \) By factor theorem, \( x + 3 \) is a factor of \( x^2 + 2x - 3 \).
In simple words: To check if \( x + 3 \) is a factor, we plug \( x = -3 \) into the polynomial. Since the final answer is 0, it means \( x + 3 \) divides the polynomial perfectly, so it is indeed a factor.

🎯 Exam Tip: State the conclusion clearly at the end. If the remainder is 0, write "Hence, it is a factor"; if not, write "Hence, it is not a factor" to secure full marks.

Question 8. If \( (x - 2) \) is a factor of \( x^3 - mx^2 + 10x - 20 \), then find the value of m.
Answer:
Let \( p(x) = x^3 - mx^2 + 10x - 20 \)
Since \( (x - 2) \) is a factor of \( p(x) \),
\( \therefore \) By factor theorem, \( p(2) = 0 \)
\( p(x) = x^3 - mx^2 + 10x - 20 \)
\( \therefore p(2) = (2)^3 - m(2)^2 + 10(2) - 20 = 0 \)
\( \therefore 8 - 4m + 20 - 20 = 0 \)
\( \therefore 8 - 4m = 0 \)
\( \therefore 4m = 8 \)
\( \therefore m = 2 \)
In simple words: Since \( x - 2 \) is a factor, plugging \( x = 2 \) into the polynomial must give a result of 0. We use this to set up an equation and solve for \( m \), which gives us 2.

🎯 Exam Tip: When a binomial is a factor, the remainder is always zero. Use this property to set up your equation directly and solve for the unknown variable.

Question 8. If \( (x-2) \) is a factor of \( p(x) = x^3 - mx^2 + 10x - 20 \), then find the value of \( m \).
Answer:
Since \( (x-2) \) is a factor of \( p(x) \), by factor theorem, the remainder is zero.
\( \therefore \text{Remainder} = p(2) = 0 \)
\( p(x) = x^3 - mx^2 + 10x - 20 \)
\( \dots p(2) = (2)^3 - m(2)^2 + 10(2) - 20 \)
\( \therefore 0 = 8 - 4m + 20 - 20 \)
\( \dots 0 = 8 - 4m \)
\( \therefore 4m = 8 \)
\( \therefore m = 2 \)
Thus, the value of \( m \) is 2.
In simple words: Since \( x-2 \) is a factor, plugging \( x = 2 \) into the polynomial must make the entire expression equal to 0. Solving this simple equation gives us \( m = 2 \).

🎯 Exam Tip: When a linear polynomial is a factor, always equate the polynomial to zero after substituting the value of \( x \) to solve for the unknown variable.

Question 9. By using factor theorem in the following examples, determine whether \( q(x) \) is a factor of \( p(x) \) or not.
(i) \( p(x) = x^3 - x^2 - x - 1 \); \( q(x) = x - 1 \)
(ii) \( p(x) = 2x^3 - x^2 - 45 \); \( q(x) = x - 3 \)
Answer:
(i) \( p(x) = x^3 - x^2 - x - 1 \)
Divisor \( = q(x) = x - 1 \)
\( \therefore \) take \( x = 1 \)
Remainder \( = p(1) \)
\( p(x) = x^3 - x^2 - x - 1 \)
\( \therefore p(1) = (1)^3 - (1)^2 - 1 - 1 \)
\( = 1 - 1 - 1 - 1 \)
\( = -2 \neq 0 \)
\( \therefore \) By factor theorem, \( x - 1 \) is not a factor of \( x^3 - x^2 - x - 1 \).

(ii) \( p(x) = 2x^3 - x^2 - 45 \)
Divisor \( = q(x) = x - 3 \)
take \( x = 3 \)
Remainder \( = p(3) \)
\( p(x) = 2x^3 - x^2 - 45 \)
\( p(3) = 2(3)^3 - (3)^2 - 45 \)
\( = 2(27) - 9 - 45 \)
\( = 54 - 9 - 45 \)
\( = 0 \)
\( \therefore \) By factor theorem, \( x - 3 \) is a factor of \( 2x^3 - x^2 - 45 \).
In simple words: To find if one expression is a factor of another, we plug the zero of the divisor into the polynomial. If the final answer is zero, it is a factor; if it is any other number, it is not.

🎯 Exam Tip: Always remember to change the sign of the constant in the divisor when substituting (e.g., for \( x - 3 \), substitute \( x = 3 \)) to avoid calculation errors.

Question 10. If \( (x^{31} + 31) \) is divided by \( (x + 1) \), then find the remainder.
Answer:
Let \( p(x) = x^{31} + 31 \)
Divisor \( = x + 1 \)
To find the remainder, take \( x = -1 \).
By Remainder Theorem:
Remainder \( = p(-1) \)
\( p(-1) = (-1)^{31} + 31 \)
Since the power is odd, \( (-1)^{31} = -1 \).
\( \therefore \text{Remainder} = -1 + 31 = 30 \)
Thus, the remainder when \( (x^{31} + 31) \) is divided by \( (x + 1) \) is 30.
In simple words: We substitute \( x = -1 \) into the expression because \( x + 1 = 0 \) gives \( x = -1 \). Since a negative number raised to an odd power stays negative, we get \( -1 + 31 \), which equals 30.

🎯 Exam Tip: Remember that any negative number raised to an odd power remains negative, whereas raised to an even power it becomes positive.

Question 11. Show that \( m - 1 \) is a factor of \( m^{21} - 1 \) and \( m^{22} - 1 \). [3 Marks]
Answer:
i. Let \( p(m) = m^{21} - 1 \)
Divisor = \( m - 1 \)
\( \therefore \) take \( m = 1 \)
By remainder theorem,
Remainder = \( p(1) \)
\( p(m) = m^{21} - 1 \)
\( \therefore p(1) = 1^{21} - 1 = 1 - 1 = 0 \)
\( \therefore \) By factor theorem, \( m - 1 \) is a factor of \( m^{21} - 1 \).

ii. Let \( p(m) = m^{22} - 1 \)
Divisor = \( m - 1 \)
\( \therefore \) take \( m = 1 \)
By remainder theorem,
Remainder = \( p(1) \)
\( p(m) = m^{22} - 1 \)
\( \therefore p(1) = 1^{22} - 1 = 1 - 1 = 0 \)
\( \dots \) By factor theorem, \( m - 1 \) is a factor of \( m^{22} - 1 \).
In simple words: To prove that \( m - 1 \) is a factor, we substitute \( m = 1 \) into both expressions. Since both calculations result in a remainder of zero, it proves that \( m - 1 \) is indeed a factor of both polynomials.

🎯 Exam Tip: Always state the Factor Theorem clearly before substituting the value, and show that the final remainder is exactly zero to secure full marks.

Question 12. If \( x - 2 \) and \( x - \frac{1}{2} \) both are the factors of the polynomial \( nx^2 - 5x + m \), then show that \( m = n = 2 \).
Answer:
Let \( p(x) = nx^2 - 5x + m \).
Since \( (x - 2) \) is a factor of \( p(x) \), by factor theorem, we have:
\( p(2) = 0 \)
\( \implies n(2)^2 - 5(2) + m = 0 \)
\( \implies 4n - 10 + m = 0 \)
\( \implies m + 4n = 10 \) — (i)

Since \( (x - \frac{1}{2}) \) is also a factor of \( p(x) \), by factor theorem, we have:
\( p(\frac{1}{2}) = 0 \)
\( \implies n(\frac{1}{2})^2 - 5(\frac{1}{2}) + m = 0 \)
\( \implies \frac{n}{4} - \frac{5}{2} + m = 0 \)
Multiplying the entire equation by 4 to clear the fractions:
\( \implies n - 10 + 4m = 0 \)
\( \implies 4m + n = 10 \) — (ii)

Equating equation (i) and equation (ii) since both are equal to 10:
\( m + 4n = 4m + n \)
\( \implies 4n - n = 4m - m \)
\( \implies 3n = 3m \)
\( \implies m = n \)

Substituting \( m = n \) in equation (i):
\( m + 4m = 10 \)
\( \implies 5m = 10 \)
\( \implies m = 2 \)
Since \( m = n \), we get \( n = 2 \).
Therefore, \( m = n = 2 \).
In simple words: Since both terms are factors, plugging in \( x = 2 \) and \( x = 1/2 \) gives us two equations equal to zero. Solving these equations together shows that both \( m \) and \( n \) must be equal to 2.

🎯 Exam Tip: When dealing with fractional factors like \( x - \frac{1}{2} \), multiply the entire equation by the denominator to eliminate fractions early and avoid calculation errors.

Question 13.
(i) If \( p(x) = 2 + 5x \), then find the value of \( p(2) + p(-2) - p(1) \).
Answer: Given polynomial is \( p(x) = 2 + 5x \). We will find the value of each term individually to compute the final expression.
For \( x = 2 \):
\( p(2) = 2 + 5(2) \)
\( = 2 + 10 \)
\( = 12 \)
For \( x = -2 \):
\( p(-2) = 2 + 5(-2) \)
\( = 2 - 10 \)
\( = -8 \)
For \( x = 1 \):
\( p(1) = 2 + 5(1) \)
\( = 2 + 5 \)
\( = 7 \)
Now, substituting these values into the expression:
\( p(2) + p(-2) - p(1) = 12 + (-8) - 7 \)
\( = 12 - 8 - 7 \)
\( = 4 - 7 \)
\( = -3 \)
In simple words: To find the final value, we substitute the numbers 2, -2, and 1 into the given formula one by one. Once we get the three individual answers, we add and subtract them as asked to get the final result of -3.

🎯 Exam Tip: Be very careful with negative signs when substituting values like \( -2 \) into the expression to avoid simple calculation errors.

Question ii. If \( p(x) = 2x^2 - 5\sqrt{3}x + 5 \), then find the value of \( p(5\sqrt{3}) \).
Answer: Given polynomial is \( p(x) = 2x^2 - 5\sqrt{3}x + 5 \).
To find \( p(5\sqrt{3}) \), we substitute \( x = 5\sqrt{3} \) into the polynomial:
\( p(5\sqrt{3}) = 2(5\sqrt{3})^2 - 5\sqrt{3}(5\sqrt{3}) + 5 \)
\( \implies p(5\sqrt{3}) = 2(25 \times 3) - (25 \times 3) + 5 \)
\( \implies p(5\sqrt{3}) = 2(75) - 75 + 5 \)
\( \implies p(5\sqrt{3}) = 150 - 75 + 5 \)
\( \implies p(5\sqrt{3}) = 75 + 5 \)
\( \implies p(5\sqrt{3}) = 80 \)
Therefore, the value of the polynomial is 80.
In simple words: To find the value, we just replace the letter \( x \) with the number \( 5\sqrt{3} \) everywhere in the expression and calculate the final basic number.

🎯 Exam Tip: When squaring a term like \( 5\sqrt{3} \), remember to square both parts: \( (5)^2 \times (\sqrt{3})^2 = 25 \times 3 = 75 \). This is a common place where students make silly calculation errors.

Question 1.
1. Divide \( p(x) = 3x^2 + x + 7 \) by \( x + 2 \). Find the remainder.
2. Find the value of \( p(x) = 3x^2 + x + 7 \) when \( x = -2 \).
3. See whether remainder obtained by division is same as the value of \( p(-2) \). Take one more example and verify.
Answer:
1. By performing polynomial long division of \( 3x^2 + x + 7 \) by \( x + 2 \):
Dividing \( 3x^2 \) by \( x \) gives \( 3x \). Multiplying \( 3x(x + 2) \) gives \( 3x^2 + 6x \). Subtracting this from \( 3x^2 + x \) leaves \( -5x \).
Next, dividing \( -5x \) by \( x \) gives \( -5 \). Multiplying \( -5(x + 2) \) gives \( -5x - 10 \). Subtracting this from \( -5x + 7 \) leaves a remainder of \( 17 \).
\( \implies \text{Remainder} = 17 \)

2. Given \( p(x) = 3x^2 + x + 7 \).
Substituting \( x = -2 \):
\( p(-2) = 3(-2)^2 + (-2) + 7 \)
\( \implies p(-2) = 3(4) - 2 + 7 \)
\( \implies p(-2) = 12 - 2 + 7 \)
\( \implies p(-2) = 17 \)

3. Yes, the remainder obtained by division is the same as the value of \( p(-2) \).

Another Example for Verification:
Let the polynomial be \( p(t) = t^3 - 3t^2 + kt + 50 \). If it is divided by \( (t - 3) \), the remainder is \( 62 \). Find the value of \( k \).
By Remainder Theorem, when \( p(t) \) is divided by \( (t - 3) \), the remainder is equal to \( p(3) \).
Given that the remainder is \( 62 \):
\( \implies p(3) = 62 \)
\( \implies 3^3 - 3(3)^2 + k(3) + 50 = 62 \)
\( \implies 27 - 27 + 3k + 50 = 62 \)
\( \implies 3k + 50 = 62 \)
\( \implies 3k = 62 - 50 \)
\( \implies 3k = 12 \)
\( \implies k = 4 \)
This confirms that the Remainder Theorem holds true for other polynomials as well.
In simple words: This exercise shows that instead of doing a long, difficult division to find the remainder, we can simply plug the opposite value of the divisor into the polynomial to get the exact same remainder instantly.

🎯 Exam Tip: The Remainder Theorem states that when a polynomial \( p(x) \) is divided by \( x - a \), the remainder is always \( p(a) \). Always use this shortcut to verify your long division answers quickly.

Question 1. Find the value of \( k \) if the remainder of the polynomial \( x^3 - 3x^2 + kx + 50 \) when divided by \( x - 3 \) is 62.
Answer:
By remainder theorem,
Remainder = \( p(3) = 3^3 - 3(3)^2 + k \times 3 + 50 \)
\( = 27 - 3 \times 9 + 3k + 50 \)
\( = 27 - 27 + 3k + 50 \)
\( = 3k + 50 \)
But the remainder is given as 62.

\( \implies 3k + 50 = 62 \)
\( \implies 3k = 62 - 50 \)
\( \implies 3k = 12 \)
\( \implies k = 4 \)
In simple words: We substitute 3 into the polynomial, set the result equal to the given remainder of 62, and solve the simple equation to find that k is 4.

🎯 Exam Tip: Always substitute the value carefully and remember to equate the final expression to the given remainder to solve for the unknown variable.

Question 2. Verify that \( (x - 1) \) is a factor of the polynomial \( x^3 + 4x - 5 \).
Answer:
Here, \( p(x) = x^3 + 4x - 5 \)
Substituting \( x = 1 \) in \( p(x) \), we get:
\( p(1) = (1)^3 + 4(1) - 5 \)
\( = 1 + 4 - 5 \)
\( = 5 - 5 \)
\( p(1) = 0 \)
By remainder theorem,
Remainder = 0
Therefore, \( (x - 1) \) is a factor of \( x^3 + 4x - 5 \).
In simple words: To check if \( (x-1) \) is a factor, we put \( x = 1 \) into the expression. Since the final answer is 0, it means \( (x-1) \) divides it perfectly with no remainder, so it is indeed a factor.

🎯 Exam Tip: To show a term is a factor, always show that the remainder \( p(a) \) equals zero. Clearly state the use of the Factor Theorem in your final line.