Question

without drawing actual graph find the zeros of the polynomials (i)x²-2x-8
(ii) -x²-2x+3
(iii) x²+x+1
(iv) x²-1
(v) x²+4x+4
(vi) -4x²+4x-1

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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

here are the following answer

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Step-by-step explanation:

(i) x2 - 4x + 2x - 8 = 0

x (x - 4) + 2 (x - 4) = 0

(x - 4) (x + 2) = 0

x = 4 , x = -2 are the zeroes of the polynomial.

Thus, α = 4, β = -2

(ii) x^2- 3x -1x -3=0

x(x+3) - 1(x+3)= 0

( x + 3) ( x - 1) = 0

x = -3, x = 1 are the zero of polynomial.

(iii) for this equation we used following formula:-

D = b^2 -4ac

= 1^2 - 4(1)(1)

= 1 - 4

= - 3

:- - 3 is smaller than 0 so..it has no real roots..

(iv) x^2-1

x^2 - 1 = 0

x^2 = 1

x = √1

x= 1 or -1

the zeros of polynomial is 1 or -1

(v) x2 + 4 x + 4

x2 + 2x + 2x +4=0

x( x + 2) +2( x + 2 )

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

here -2 & -2 are zero of the polynomial

vi 4x2+4x+1

=4x2+2x+2x+1, (split middle term )

=(4x2+2x)+(2x+1)

=2x(2x+1)+1(2x+1),

=(2x+1)(2x+1),

factor (2x+1)=0 or (2x+1) =0

zeros of polynomial is 1/2 or 1/2