RBSE Solutions Class 9 Maths Chapter 3 Polynomial Exercise 3.2

Polynomial Ex 3.2

Question 1. Find the value of the polynomial \( 2x^3 - 13x^2 + 17x + 12 \) at
(i) \( x = 2 \)
(ii) \( x = -3 \)
(iii) \( x = 0 \)
(iv) \( x = -1 \)
Answer: Let the polynomial be \( p(x) = 2x^3 - 13x^2 + 17x + 12 \).
(i) For \( x = 2 \):
\( p(2) = 2(2)^3 - 13(2)^2 + 17(2) + 12 \)
\( = 2(8) - 13(4) + 34 + 12 \)
\( = 16 - 52 + 34 + 12 \)
\( = 62 - 52 \)
\( = 10 \)
(ii) For \( x = -3 \):
\( p(-3) = 2(-3)^3 - 13(-3)^2 + 17(-3) + 12 \)
\( = 2(-27) - 13(9) - 51 + 12 \)
\( = -54 - 117 - 51 + 12 \)
\( = -222 + 12 \)
\( = -210 \)
(iii) For \( x = 0 \):
\( p(0) = 2(0)^3 - 13(0)^2 + 17(0) + 12 \)
\( = 0 - 0 + 0 + 12 \)
\( = 12 \)
(iv) For \( x = -1 \):
\( p(-1) = 2(-1)^3 - 13(-1)^2 + 17(-1) + 12 \)
\( = 2(-1) - 13(1) - 17 + 12 \)
\( = -2 - 13 - 17 + 12 \)
\( = -32 + 12 \)
\( = -20 \)
Evaluating polynomials at specific points is crucial in many areas of mathematics, like graphing and finding roots.
In simple words: We put the given 'x' numbers into the polynomial equation. Then, we do the math step by step to find the final answer for each 'x'.

🎯 Exam Tip: Be very careful with signs (positive and negative) when substituting negative values for x, especially with powers.

Question 2. Find \( p(2) \), \( p(1) \) and \( p(0) \) for each of the following polynomials.
(i) \( p(x) = x^2 - x + 1 \)
(ii) \( p(y) = (y + 1)(y - 1) \)
(iii) \( p(x) = x^3 \)
(iv) \( p(t) = 2 + t + t^2 - t^3 \)
Answer:
(i) For \( p(x) = x^2 - x + 1 \):
For \( x = 2 \): \( p(2) = (2)^2 - (2) + 1 = 4 - 2 + 1 = 3 \)
For \( x = 1 \): \( p(1) = (1)^2 - (1) + 1 = 1 - 1 + 1 = 1 \)
For \( x = 0 \): \( p(0) = (0)^2 - (0) + 1 = 0 - 0 + 1 = 1 \)
(ii) For \( p(y) = (y + 1)(y - 1) \):
For \( y = 2 \): \( p(2) = (2 + 1)(2 - 1) = (3)(1) = 3 \)
For \( y = 1 \): \( p(1) = (1 + 1)(1 - 1) = (2)(0) = 0 \)
For \( y = 0 \): \( p(0) = (0 + 1)(0 - 1) = (1)(-1) = -1 \)
(iii) For \( p(x) = x^3 \):
For \( x = 2 \): \( p(2) = (2)^3 = 8 \)
For \( x = 1 \): \( p(1) = (1)^3 = 1 \)
For \( x = 0 \): \( p(0) = (0)^3 = 0 \)
(iv) For \( p(t) = 2 + t + t^2 - t^3 \):
For \( t = 2 \): \( p(2) = 2 + 2 + (2)^2 - (2)^3 = 2 + 2 + 4 - 8 = 8 - 8 = 0 \)
For \( t = 1 \): \( p(1) = 2 + 1 + (1)^2 - (1)^3 = 2 + 1 + 1 - 1 = 3 \)
For \( t = 0 \): \( p(0) = 2 + 0 + (0)^2 - (0)^3 = 2 + 0 + 0 - 0 = 2 \)
Finding the value of a polynomial at specific points helps us understand its behavior, such as where it crosses the x-axis or its highest/lowest points.
In simple words: For each polynomial given, we replaced the letter (like x, y, or t) with 2, then 1, and then 0. We did the calculations to find what number the polynomial became each time.

🎯 Exam Tip: Remember to substitute the value into every term where the variable appears in the polynomial.

Question 3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) \( p(x) = x^2 - 1; x = 1, -1 \)
(ii) \( p(x) = 2x + 1; x = -\frac{1}{2} \)
(iii) \( p(x) = 4x + 5; x = -\frac{5}{4} \)
(iv) \( p(x) = 3x^2; x = 0 \)
(v) \( p(x) = (x - 3)(x + 5); x = 3, -5 \)
(vi) \( p(x) = ax + b; x = -\frac{b}{a} \)
(vii) \( p(x) = 3x^2 - 1; x = -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \)
(viii) \( p(x) = 3x + 2; x = -\frac{2}{3} \)
Answer: A value of \( x \) is a zero of a polynomial \( p(x) \) if \( p(x) = 0 \) when that value is substituted.
(i) For \( p(x) = x^2 - 1 \):
For \( x = 1 \): \( p(1) = (1)^2 - 1 = 1 - 1 = 0 \)
For \( x = -1 \): \( p(-1) = (-1)^2 - 1 = 1 - 1 = 0 \)
Yes, \( 1 \) and \( -1 \) are zeroes of the polynomial \( p(x) = x^2 - 1 \).
(ii) For \( p(x) = 2x + 1 \):
For \( x = -\frac{1}{2} \): \( p(-\frac{1}{2}) = 2(-\frac{1}{2}) + 1 = -1 + 1 = 0 \)
Yes, \( -\frac{1}{2} \) is a zero of the polynomial \( p(x) = 2x + 1 \).
(iii) For \( p(x) = 4x + 5 \):
For \( x = -\frac{5}{4} \): \( p(-\frac{5}{4}) = 4(-\frac{5}{4}) + 5 = -5 + 5 = 0 \)
Yes, \( -\frac{5}{4} \) is a zero of the polynomial \( p(x) = 4x + 5 \).
(iv) For \( p(x) = 3x^2 \):
For \( x = 0 \): \( p(0) = 3(0)^2 = 3(0) = 0 \)
Yes, \( 0 \) is a zero of the polynomial \( p(x) = 3x^2 \).
(v) For \( p(x) = (x - 3)(x + 5) \):
For \( x = 3 \): \( p(3) = (3 - 3)(3 + 5) = (0)(8) = 0 \)
For \( x = -5 \): \( p(-5) = (-5 - 3)(-5 + 5) = (-8)(0) = 0 \)
Yes, both \( 3 \) and \( -5 \) are zeroes of the polynomial \( p(x) = (x - 3)(x + 5) \).
(vi) For \( p(x) = ax + b \):
For \( x = -\frac{b}{a} \): \( p(-\frac{b}{a}) = a(-\frac{b}{a}) + b = -b + b = 0 \)
Yes, \( -\frac{b}{a} \) is a zero of the polynomial \( p(x) = ax + b \).
(vii) For \( p(x) = 3x^2 - 1 \):
For \( x = -\frac{1}{\sqrt{3}} \): \( p(-\frac{1}{\sqrt{3}}) = 3(-\frac{1}{\sqrt{3}})^2 - 1 = 3(\frac{1}{3}) - 1 = 1 - 1 = 0 \)
For \( x = \frac{1}{\sqrt{3}} \): \( p(\frac{1}{\sqrt{3}}) = 3(\frac{1}{\sqrt{3}})^2 - 1 = 3(\frac{1}{3}) - 1 = 1 - 1 = 0 \)
Yes, both \( -\frac{1}{\sqrt{3}} \) and \( \frac{1}{\sqrt{3}} \) are zeroes of the polynomial \( p(x) = 3x^2 - 1 \).
(viii) For \( p(x) = 3x + 2 \):
For \( x = -\frac{2}{3} \): \( p(-\frac{2}{3}) = 3(-\frac{2}{3}) + 2 = -2 + 2 = 0 \)
Yes, \( -\frac{2}{3} \) is a zero of the polynomial \( p(x) = 3x + 2 \).
Finding the zeroes of a polynomial is fundamental because they represent the x-intercepts of the polynomial's graph, where the function's value is zero.
In simple words: We check if a given number makes the polynomial equal to zero. If it does, then that number is called a "zero" of the polynomial. We replace the variable with the given number and do the math to see if the answer is zero.

🎯 Exam Tip: A zero of a polynomial means that when you substitute that value into the polynomial, the result is exactly zero. This is a key concept for solving equations.

Question 4. Find the zeroes of the polynomial in each case.
(i) \( p(x) = x - 4 \)
(ii) \( p(x) = 4x \)
(iii) \( p(x) = bx, b \neq 0 \)
(iv) \( p(x) = x + 3 \)
(v) \( p(x) = 2x - 1 \)
(vi) \( p(x) = 3x + 7 \)
(vii) \( p(x) = cx + d, c \neq 0, c \) and \( d \) are real numbers.
Answer: To find the zeroes of a polynomial, we set the polynomial equal to zero and solve for the variable.
(i) For \( p(x) = x - 4 \):
Set \( p(x) = 0 \)
\( \implies x - 4 = 0 \)
\( \implies x = 4 \)
The zero of the polynomial is \( 4 \).
(ii) For \( p(x) = 4x \):
Set \( p(x) = 0 \)
\( \implies 4x = 0 \)
\( \implies x = \frac{0}{4} \)
\( \implies x = 0 \)
The zero of the polynomial is \( 0 \).
(iii) For \( p(x) = bx, b \neq 0 \):
Set \( p(x) = 0 \)
\( \implies bx = 0 \)
\( \implies x = \frac{0}{b} \) (since \( b \neq 0 \))
\( \implies x = 0 \)
The zero of the polynomial is \( 0 \).
(iv) For \( p(x) = x + 3 \):
Set \( p(x) = 0 \)
\( \implies x + 3 = 0 \)
\( \implies x = -3 \)
The zero of the polynomial is \( -3 \).
(v) For \( p(x) = 2x - 1 \):
Set \( p(x) = 0 \)
\( \implies 2x - 1 = 0 \)
\( \implies 2x = 1 \)
\( \implies x = \frac{1}{2} \)
The zero of the polynomial is \( \frac{1}{2} \).
(vi) For \( p(x) = 3x + 7 \):
Set \( p(x) = 0 \)
\( \implies 3x + 7 = 0 \)
\( \implies 3x = -7 \)
\( \implies x = -\frac{7}{3} \)
The zero of the polynomial is \( -\frac{7}{3} \).
(vii) For \( p(x) = cx + d, c \neq 0 \):
Set \( p(x) = 0 \)
\( \implies cx + d = 0 \)
\( \implies cx = -d \)
\( \implies x = -\frac{d}{c} \) (since \( c \neq 0 \))
The zero of the polynomial is \( -\frac{d}{c} \).
Finding zeroes helps us solve equations and determine where a function crosses or touches the x-axis, which is essential for graphing and understanding function behavior.
In simple words: To find the "zero" of a polynomial, we make the polynomial equal to zero. Then, we solve that simple equation to find the value of the variable. This value is the zero.

🎯 Exam Tip: Remember that for a linear polynomial like \( ax+b \), the zero can always be found directly by setting the expression to zero and solving for \( x \), which gives \( x = -\frac{b}{a} \).