Question

Find the zeroes of x2

– x – 12 graphically.​

Answer 1

Given polynomial p(x) = x² - x - 12

Let y = x² - x - 12

Plotting the points on a graph and joining them with a smooth curve we get a parabola.

The parabola meets the x - axis at ( -3 , 0 ) and

(4,0) .

So, the Zeroes are -3 and 4.

Verification:

$x^{2} - x - 12 = 0$

/* Splitting the middle term, we get */

$\implies x^{2} - 4x + 3x - 12 = 0$

$\implies x( x - 4) + 3( x - 4 ) = 0$

$\implies (x-4)(x+3) = 0$

$\implies x - 4 = 0 \:Or \: x + 3 = 0$

$\implies x = 4 \:Or \: x = -3$

$\therefore -3\: and \: 4 \: are \: zeroes \: of$

$the \: polynomial.$

•••♪