Question

Find the zeros of the given polynomials.
(i) p(x) = 3x (ii) p(x) = x² + 5x + 6
(iii) p(x) = (x+2) (x+3) (iv) p(x) = x4 – 16

Answer 1

Dear Student:

i) p(x) = 3x

p(x)=0

3x=0

x=0

(ii) p(x) = x² + 5x + 6

p(x)=0

x² + 5x + 6 =0

x² + 2x +3x+ 6 =0

x(x+2)+3(x+2)=0

(x+3)(x+2)=0

(x+3)=0

x=-3

(x+2)=0

x=-2

(iii) p(x) = (x+2) (x+3)

(x+3)(x+2)=0

(x+3)=0

x=-3

(x+2)=0

x=-2

(iv) p(x) = x^4 – 16

$a^{4} -b^{4} =(a^{2}-b^{2})(a^{2}+b^{2})$

a=x and b=2

$x^{4} -2^{4} =(x^{2}-2^{2})(x^{2}+2^{2})=0$

$(x^{2}-2^{2})=0$

x=2 and -2

$(x^{2}+2^{2})=0$

x=-2i where i is iota.

Answer 2

Solution :

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We get zeroes of the polynomial

p(x) we will take p(x) = 0

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i ) p(x) = 3x

p(x) = 0

=> 3x = 0

=> x = 0

The zero of p(x) is 0.

ii ) p(x) = x² + 5x + 6

Let p(x) = 0

=> x² + 5x + 6 = 0

=> x² + 2x + 3x + 6 = 0

=> x( x + 2 ) + 3( x + 2 ) = 0

=> ( x + 2 )( x + 3 ) = 0

=> x + 2 = 0 or x + 3 = 0

=> x = -2 or x = -3

The zeroes of given polynomial are

-2 , -3

iii ) p(x) = ( x + 2 )( x + 3 )

Let p(x) = 0

=> (x+2)(x+3) = 0

=> x + 2 = 0 or x + 3 = 0

=> x = -2 or x = -3

Therefore ,

The Zeroes of given polynomial

are -2 , -3

iv ) p(x) = x⁴ - 16

Let p(x) = 0

=> x⁴ - 16 = 0

=> ( x² )² - 4² = 0

=> ( x² + 4 )( x² - 4 ) = 0

=> ( x² + 4 )( x² - 2² ) = 0

=> ( x² + 4 ) ( x + 2 )( x - 2 ) = 0

=> x² + 4 = 0 or x+2 = 0 or x-2 = 0

=> x² = -4 or x = -2 or x = 2

=> x = ± 2i , x = -2 , x = 2

Therefore ,

The zeroes of the given polynomial

are ± 2i , -2 , 2

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