NCERT solution class 9 chapter 2 Polynomials exercise 2.3 mathematics

EXERCISE 2.3

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Q 1 (iv , v and vi), Ex 2.3, Page No 40, Polynomials, Maths CBSE Class 9th

Page No 34:

Question 1:

Find the value of the polynomial $5 x-4 x^{2}+3$ at
(i) x = 0 (ii) x = −1 (iii) x = 2

Answer:

(i) $p(x)=5 x-4 x^{2}+3$
$p(0)=5(0)-4(0)^{2}+3$
=3

(ii) $p(x)=5 x-4 x^{2}+3$
$p(-1)=5(-1)-4(-1)^{2}+3$
=-5-4(1)+3
=-6

(iii) $p(x)=5 x-4 x^{2}+3$
$p(2)=5(2)-4(2)^{2}+3$
=10-16+3
=-3

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Q 2, Ex 2.3, Page No 40, Polynomials, Class 9th Mathematics

Question 2:

Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y² − y + 1
(ii) p(t) = 2 + t + 2t² − t³
(iii) p(x) = x³ (iv) p(x) = (x − 1) (x + 1)

Answer:

(i) p(y) = y² − y + 1
p(0) = (0)² − (0) + 1 = 1
p(1) = (1)² − (1) + 1 = 1
p(2) = (2)² − (2) + 1 = 3

(ii) p(t) = 2 + t + 2t² − t³
p(0) = 2 + 0 + 2 (0)² − (0)³ = 2
p(1) = 2 + (1) + 2(1)² − (1)³
= 2 + 1 + 2 − 1 = 4
p(2) = 2 + 2 + 2(2)² − (2)³
= 2 + 2 + 8 − 8 = 4

(iii) p(x) = x³
p(0) = (0)³ = 0
p(1) = (1)³ = 1
p(2) = (2)³ = 8
(iv) p(x) = (x − 1) (x + 1)
p(0) = (0 − 1) (0 + 1) = (− 1) (1) = − 1
p(1) = (1 − 1) (1 + 1) = 0 (2) = 0
p(2) = (2 − 1 ) (2 + 1) = 1(3) = 3

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Q 3, Ex 2.3, Page No 40, Polynomials, Maths CBSE Class 9th NCERT

Page No 35:

Question 3:

Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x)=3x+1, $x=-\frac{1}{3}$
(ii) p(x)=5x-π , $x=\frac{4}{5}$
(iii) p(x) = x² − 1, x = 1, − 1
(iv) p(x) = (x + 1) (x − 2), x = − 1, 2
(v) p(x) = x² , x = 0 (vi) p(x) = lx + m
(vii) $p(x)=3 x^{2}-1$,  $x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$
(viii) p(x)=2x+1,  $x=\frac{1}{2}$

Answer:

(i) If $x=\frac{-1}{3}$ is a zero of given polynomial p(x) = 3x + 1, then $p\left(-\frac{1}{3}\right)$ should be 0.
Here, $p\left(\frac{-1}{3}\right)=3\left(\frac{-1}{3}\right)+1$

=-1+1=0

Therefore, $x=\frac{-1}{3}$ is a zero of the given polynomial.

(ii) If  $x=\frac{4}{5}$ is a zero of polynomial p(x) = 5x − π , then $p\left(\frac{4}{5}\right)$ should be 0.

Here , $p\left(\frac{4}{5}\right)=5\left(\frac{4}{5}\right)-\pi$
=4-π

As $p\left(\frac{4}{5}\right) \neq 0$

Therefore, $x=\frac{4}{5}$ is not a zero of the given polynomial.


(iii) If x = 1 and x = −1 are zeroes of polynomial p(x) = x² − 1, then p(1) and p(−1) should be 0.
Here, p(1) = (1)² − 1 = 0, and
p(− 1) = (− 1)² − 1 = 0
Hence, x = 1 and −1 are zeroes of the given polynomial.


(iv) If x = −1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x − 2), then p(−1) and p(2)should be 0.
Here, p(−1) = (− 1 + 1) (− 1 − 2) = 0 (−3) = 0, and
p(2) = (2 + 1) (2 − 2 ) = 3 (0) = 0
Therefore, x = −1 and x = 2 are zeroes of the given polynomial.


(v) If x = 0 is a zero of polynomial p(x) = x², then p(0) should be zero.
Here, p(0) = (0)² = 0
Hence, x = 0 is a zero of the given polynomial.


(vi) If $x=\frac{-m}{l}$ is a zero of polynomial p(x) = lx + m, then $p\left(\frac{-m}{l}\right)$ should be 0.
Here,$p\left(\frac{-m}{l}\right)=l\left(\frac{-m}{l}\right)+m$
=-m+m=0
Therefore, $x=-\frac{m}{l}$ is a zero of the given polynomial.


(vii) If  $x=\frac{-1}{\sqrt{3}}$and $x=\frac{2}{\sqrt{3}}$ are zeroes of polynomial p(x) = 3x² − 1, then
$p\left(\frac{-1}{\sqrt{3}}\right)$ and $p\left(\frac{2}{\sqrt{3}}\right)$ should be 0

Here , $p\left(\frac{-1}{\sqrt{3}}\right)=3\left(\frac{-1}{\sqrt{3}}\right)^{2}-1$
$=3\left(\frac{1}{3}\right)-1$
=1-1
=0

and

$p\left(\frac{2}{\sqrt{3}}\right)=3\left(\frac{2}{\sqrt{3}}\right)^{2}-1$
$=3\left(\frac{4}{3}\right)-1$
=4-1
=3

Hence, $x=\frac{-1}{\sqrt{3}}$ is a zero of the given polynomial. However, $x=\frac{2}{\sqrt{3}}$ is not a zero of the given polynomial.


(viii) If $x=\frac{1}{2}$ is a zero of polynomial p(x) = 2x + 1, then $p\left(\frac{1}{2}\right)$ should be 0.

Here , $p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)+1$
=1+1
=2

As $p\left(\frac{1}{2}\right) \neq 0$ ,

Therefore, $x=\frac{1}{2}$ is not a zero of the given polynomial.

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Question 4:

Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5 (ii) p(x) = x − 5 (iii) p(x) = 2x + 5
(iv) p(x) = 3x − 2 (v) p(x) = 3x (vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0, c, are real numbers.

Answer:

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.
(i) p(x) = x + 5
p(x) = 0
x + 5 = 0
x = − 5
Therefore, for x = −5, the value of the polynomial is 0 and hence, x = −5 is a zero of the given polynomial.

(ii) p(x) = x − 5
p(x) = 0
x − 5 = 0
x = 5
Therefore, for x = 5, the value of the polynomial is 0 and hence, x = 5 is a zero of the given polynomial.

(iii) p(x) = 2x + 5
p(x) = 0
2x + 5 = 0
2x = − 5
$x=-\frac{5}{2}$
Therefore, for$x=-\frac{5}{2}$, the value of the polynomial is 0 and hence, $x=-\frac{5}{2}$ is a zero of the given polynomial.

(iv) p(x) = 3x − 2
p(x) = 0
3x − 2 = 0
$x=\frac{2}{3}$
Therefore, for $x=\frac{2}{3}$, the value of the polynomial is 0 and hence, $x=\frac{2}{3}$ is a zero of the given polynomial.

(v) p(x) = 3x
p(x) = 0
3x = 0
x = 0
Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.

(vi) p(x) = ax
p(x) = 0
ax = 0
x = 0
Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.

(vii) p(x) = cx + d
p(x) = 0
cx+ d = 0
$x=\frac{-d}{c}$

Therefore, for $x=\frac{-d}{c}$ , the value of the polynomial is 0 and hence, $x=\frac{-d}{c}$ is a zero of the given polynomial.

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