Question

What are the real zeroes

Answer 1

ANSWER

B. -3,3,-6

EXPLANATION

We equate the given polynomial to zero.

${x}^{3} + 6 {x}^{2} - 9x - 54 = 0$

Factor by grouping;

${x}^{2} (x + 6) - 9(x + 6) = 0$

Factor further;

$( {x}^{2} - 9)(x + 6) = 0$

Either

${x}^{2} - 9 = 0 \: or \: x + 6 = 0$

${x}^{2} = 9 \: or \: x = - 6$

$x = \pm \sqrt{9} \: or \: x = - 6$

$x = \pm 3 \: or \: x = - 6$

The zeroes are

x=3,-3,-6

Answer 2

You have to plug the values in the expression and see if it returns zero.

The only points where the function evaluates to zero are $x=\pm 3$ and $x=6$