Polynomial 2.2 class 9th

Ex 2.2 Class 9 Maths Question 1.
Find the value of the polynomial 5x – 4x² + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2

Solution:

(i) x = 0

5x – 4x² + 3

p(0) = 5(0) – 4(0)²+ 3

p(0) = 0 -0 + 3

p(0) = 3

(ii) x = – 1

5x – 4x² + 3

p(-1) = 5(-1) – 4(-1)² + 3

p(-1) = -5 – 4 + 3

p(-1) = -9 + 3

p(-1) = -6

(iii) x = 2

5x – 4x² + 3

p(2) = 5(2) – 4(2)² + 3

p(2) = 10 – 4(4) + 3

p(2) = 10 – 16 + 3

p(2) = -6 + 3

p(2) = -3

Ex 2.2 Class 9 Maths 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)

Solution:

p = 0

(i) p(y) = y²– y + 1

p(0) = (0)² – (0) + 1

p(0) = 0 – 0 + 1

p(0) = 1

p = 1

(i) p(y) = y²– y + 1

p(1) = (1)² – (1) + 1

p(1) = 1 – 1 + 1

p(1) = 1

p = 2

(i) p(y) = y²– y + 1

p(2) = (2)² – (2) + 1

p(2) = 4 – 2 + 1

p(2) = 2 + 1

p(2) = 3

p = 0

(ii) p(t) = 2 + t + 2t²– t³

p(0) = 2 + 0 + 2(0)² – (0)³

p(0) = 2 + 0 + 0 – 0

p(0) = 2

p = 1

(ii) p(t) = 2 + t + 2t²– t³

p(1) = 2 + (1) + 2(1)² – (1)³

p(1) = 2 + 1 + 2 – 1

p(1) = 5 – 1

p(1) = 4

p = 2

(ii) p(t) = 2 + t + 2t²– t³

p(2) = 2 + 2 + 2(2)² – (2)³

p(2) = 2 + 2 + 2(4) -(8)

p(2) = 4 + 8 – 8

p(2) = 4

p = 0

(iii) p(x) = x³

p(0) = (0)³

p(0) = 0

p = 1

(iii) p(x) = x³

p(1) = (1)³

p(1) = 1

p = 2

(iii) p(x) = x³

p(2) = (2)³

p(2) = 8

p = 0

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

p(0) = ( 0 – 1 ) ( 0 + 1 )

p(0) = ( -1 ) ( 1 )

p(0) = -1

p = 1

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

p(1) = ( 1 – 1 ) ( 1 + 1 )

p(1) = ( 0 ) ( 2 )

p(1) = 0

p = 2

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

p(2) = ( 2 – 1 ) ( 2 + 1 )

p(2) = ( 1 ) ( 3 )

p(2) = 3

Ex 2.2 Class 9 Maths 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 , x = – $\frac{m}{l}$
(vii) p(x) = 3x²-1 , x = – $\frac{1}{\sqrt{3}}$ , $\frac{2}{\sqrt{3}}$
(viii) p(x) = 2x + 1 , x = $\frac{1}{2}$

Solution:

(i) p(x) = 3x + 1 , x = – $\frac{1}{3}$

p(- $\frac{1}{3}$ ) = 3 ( – $\frac{1}{3}$ ) + 1

= $- \frac{3}{3}$ + 1

= – 1 + 1

= 0

– $\frac{1}{3}$ is a zeros of p(x)

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

p( $\frac{4}{5}$ ) = 5 ( $\frac{4}{5}$ ) – π

= $\frac{20}{5}$ – π

= 4 – π

x = $\frac{4}{5}$ is not zeros of p(x)

(iii) p(x) = x² – 1 , x = 1 , – 1

p(x) = x² – 1

p( 1 ) = 1² – 1

= 1 – 1

= 0

x = 1 is a zeros of p(x)

p(x) = x² – 1

p(-1) = -1² – 1

= 1 – 1

= 0

x = – 1 is a zeros of p(x)

(iv) p(x) = (x+1) (x-2) , x = – 1 , 2

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

p(-1) = (-1 + 1) (-1 – 2)

= 0 ( – 3 )

= 0

x = – 1 is a zeros of p(x)

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

p(2) = (2 + 1) (2 – 2)

= ( 3 ) ( 0 )

= 0

x = 2 is a zeros of p(x)

(v) p(x) = x², x = 0

p(x) = x²

p(0) = (0)²

= 0

x = 0 is a zeros of p(x)

(vi) p(x) = lx + m , x = – $\frac{m}{l}$

p(x) = lx + m

p(– $\frac{m}{l}$ ) = l (– $\frac{m}{l}$ ) + m

= – $\frac{l\times m}{l}$ + m

= – m + m

= 0

x = – $\frac{m}{l}$ is a zeros of p(x)

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

p(x) = 3x²-1

p(– $\frac{1}{\sqrt{3}}$ ) = 3(– $\frac{1}{\sqrt{3}}$ )² – 1

= $\frac{3\times 1}{3}$ – 1

= 1 – 1

= 0

x = – $\frac{1}{\sqrt{3}}$ is a zeros of p(x)

p(x) = 3x²-1

p( $\frac{2}{\sqrt{3}}$ ) = 3( $\frac{2}{\sqrt{3}}$ )² – 1

= $\frac{3\times 4}{3}$ – 1

= 4 – 1

= 3 ≠ 0

x = $\frac{2}{\sqrt{3}}$ is not a zeros of p(x)

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

p(x) = 2( $\frac{1}{2}$ ) + 1

= $\frac{2\times 1}{2}$ + 1

= 1 + 1

= 2 ≠ 0

x = $\frac{1}{2}$ is not a zeros of p(x)

Ex 2.2 Class 9 Maths 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
(vi) p(x) = 3x – 2
(v) p(x) = 3x

Solution:

(i) p(x) = x + 5

now p(x) = 0

then , x + 5 = 0

x = – 5

$\therefore$ – 5 is a zero of the polynomial p(x).