Question

Find the zero's of the polynomial x2-2x-8​

Answer 1

$\huge\tt{\red{\underline{Given:}}}$

★$A\:polynomial\:x^{2}-2x-8$ [say p(x) ].

$\huge\tt{\red{\underline{To\:\:Find:}}}$

★The zeroes of the polynimial.

$\huge\tt{\red{\underline{Concept\:\:Used:}}}$

★We can find it's zeroes either by factorising using middle term splitting or we can find it's zeroes directly using the standard formula.

i. e. ${\underline{\boxed{\red{\dfrac{-b±\sqrt{b^{2}-4ac}}{2a}}}}}$ of the equation in $ax^{2}+bx+c$ form.

$\huge\tt{\red{\underline{Answer:}}}$

We have,

$\implies p(x) =x^{2}-2x-8$

On equating with zero,

$\implies x^{2}-2x-8=0$

$\implies x^{2}-4x+2x-8=0$

$\implies x(x-4) +2(x-4) =0$

$\implies (x+2) (x-4) =0$

${\underline{\boxed{\purple{x=-2, 4}}}}$

Therefore the zeroes are -2, 4.

Answer 2

Step-by-step explanation:

$x ^{2} - 2x - 8$

$x ^{2} - 4x + 2x - 8$

$x(x - 4) + 2(x - 4)$

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

$x = 4 \: and \: - 2$