Question

Q1)Find the zeroes of the following quadratic polynomials.[06marks]

1)x^2-2x-8

Answer 1

Step-by-step explanation:

x2-2x-8=0

x2-4x+2x-8=0

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

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

x+2=0 or x-4=0

x=-2. or. x=4

hope helpful

Answer 2

Answer:

Quadratic polynomial -

$\\ \tt \: {x}^{2} - 2x - 8 = {x}^{2} - 4x + 2x - 8 \\ \\ \\ \implies \tt \: x \: (x - 4) + 2 \: (x - 4) \\ \\ \\ \implies \tt \blue{(x + 2) \: (x - 4)}$

Here, zeroes are -2 and 4 because value of x² - 2x - 8 is zero

$\\ \qquad \quad \tt \: x + 2 = 0 \\ \\ \qquad \implies \tt \: x = - 2$

or,

$\\ \qquad \quad \tt \: x - 4 = 0 \\ \\ \qquad \implies \tt \: x = 4$