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Execise. 2.2

1. Find the value of the polynomial 5x – 4x² +3 at
(i) x = 0
(ii) x = -1
(iii) x = 2

Sol.   Let f (x) = 5x – 4x²  + 3.
(i) f (0) = 0 – 0 + 3 = 3
(ii) f (-1) = – 5 – 4 + 3 = – 6
(iii) f (2) = 10 – 16 + 3 = – 3

2. Find p(0), p(1) for each of the following polynomials:
(i) p(y) = y² – y + 1

Ans. – p(0) = 0 – 0 + 1 = 1 ; p (1) = 1 – 1 + 1 = 1;
p (2) = 4 – 2 + 1 = 3.

(ii) p(t) = 2 + t + 2t² – t³

Ans. – p(0) = 2 + 0 + 0 – 0 = 2;     p(1) = 2 + 1 + 2  – 1 = 4;
p (2) = 2 + 2 + 8 – 8 = 4

(iii) p (x)  = x³

Ans. – p (0) = 0;   p(1) = 1;    p(2) =  8

(iv) p(x) = (x-1)(x+1)

Ans. – p(0) = (0 – 1) (0 + 1 ) = – 1 ;     p (1) = (1-1)(1+1) = 0 ;
p(2) = ( 2-1) (2 + 1) = 3

3. Verify whether the following are zeroes of the polynomial, indicated again indicated against them.
(i) p(x) = 3x + 1, x = $- {1 \over 3}$

Ans.-  $p\left( { - {1 \over 3}} \right) = 3 \times \left( { - {1 \over 3}} \right) + 1 = - 1 + 1 = 0$
Hence, $x = - {1 \over 3}$ is a zero of the polynomial p(x).

(ii) p(x) = 5x – π , x = ${4 \over 5}$

Ans.- $p\left( {{4 \over 5}} \right) = 5 \times {4 \over 5} - \pi = 4 - \pi \ne 0$
Hence, $x = {4 \over 5}$ is not a zero of the polynomial p (x)

(iii) p(x) =x2 – 1 , x = 1 , -1

Ans.-  p(1) = 1-1 = 0 and p(-1) = 1 -1 = 0
Hence , x = 1 and x = -1 are zeroes of the polynomial p(x)

(iv) p(x) = (x +1 ) (x-2) , x = -1 , 2

Ans.-  p(-1) = (-1 + 1 ) (-1-2) = 0  and p(2) = (2+1) (2-2) = 0
Hence, x = -1 and x=2 are zeroes of the polynomial p(x).

(v) p(x) = x², x= 0

Ans.- p(0) = 0 . Hence, x = 0 is a zero of the polynomial p (x).

(vi) p(x) = lx + m, x = $- {m \over t}$

Ans.- $p\left( { - {m \over l}} \right) = l.\left( { - {m \over l}} \right) + m = - m + m = 0$
Hence, x = –${m \over l}$ is  a zero of the polynomial p(x) .

(vii)  p(x) = 3x² – 1, x = –${1 \over {\sqrt 3 }},{2 \over {\sqrt 3 }}$

Ans.-  $p\left( { - {1 \over {\sqrt 3 }}} \right) = 3 \times {1 \over 3} - 1 = 1 - 1 = 0$
and            $p\left( {{2 \over {\sqrt 3 }}} \right) = 3 \times {4 \over 3} - 1 = 4 - 1 = 3 \ne 0$
Hence, $x = - {1 \over {\sqrt 3 }}$ is a zero and  $x = {2 \over {\sqrt 3 }}$ is not a zero of the polynomial p(x)

(viii) p(x) = 2x + 1,  x= ${1 \over 2}$

Ans.- $p\left( {{1 \over 2}} \right) = 2 \times {1 \over 2} + 1 = 1 + 1 = 2 \ne 0$
Hence, $x = {1 \over 2}$ is not a zero of the polynomial p(x).

4. Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5

Ans. –  For zero, p(x) = 0 ⇒ x + 5 = 0
⇒ x = -5 is a zero of the polynomial p(x).

(ii) p(x) = x – 5

Ans. – For zero , p(x) = 0  ⇒ x – 5 = 0
⇒ x = 5 is a zero of the polynomial p(x).

(iii) p(x) = 2x + 5

Ans. – For zero, p(x) = 0 ⇒ 2x = 5 = 0
⇒ $x = - {5 \over 2}$ is a zero of the polynomial p(x).

(iv) p(x) = 3x – 2

Ans. –  For zero, p(x)  = 0  ⇒   3x  – 2 = 0
$\Rightarrow x = {2 \over 3}$ is a zero of the polynomial p(x).

(v) p(x) = 3x

Ans. –   For zero , p(x) = 0 ⇒ 3x = 0
⇒ x = 0 is a zero of the polynomial p(x).

(vi) p(x) = ax, a ≠ 0

Ans. – For zero , p(x) = 0 ⇒ ax = 0
⇒  x = 0 , as a ≠ 0
Therefore, x = 0 is a zero of the polynomial p (x).

(vii) p(x) = cx + d, c≠0, c, d are real numbers.

$\Rightarrow x = - {d \over c},(c \ne 0)$
Therefore, $x = - {d \over c}$  is a zero of the polynomial p(x).