are the factors of the polynomial $x^{2} + 2 x - 8$ .

(x - 4) & (x - 2)

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(x + 4) & (x - 2)

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(x - 4) & (x + 2)

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Solution

The correct option is B $(x + 4) & (x - 2)$
Lets factorise $x^{2} + 2 x - 8$ :
We need to find $a$ and $b$ such that $a + b = 2$ and $a \times b = - 8$ .
We see that, $4 + (- 2) = 2$ and $4 \times - 2 = - 8$
$x^{2} + 2 x - 8 = x^{2} + 4 x - 2 x - 8 = x (x + 4) - 2 (x + 4) = (x + 4) (x - 2)$
$∴ (x + 4) & (x - 2)$ are the factors of the polynomial $x^{2} + 2 x - 8$

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are the factors of the polynomial x2+2x−8.

The correct option is B (x+4) & (x−2)Lets factorise x2+2x−8: We need to find a and b such that a+b=2 and a×b=−8. We see that, 4+(−2)=2 and 4×−2=−8 x2+2x−8=x2+4x−2x−8=x(x+4)−2(x+4)=(x+4)(x−2) ∴(x+4) & (x−2) are the factors of the polynomial x2+2x−8

are the factors of the polynomial $x^{2} + 2 x - 8$ .