Question

Please help?! Which of the following represents the zeros of f(x)=5x^3-6x^2-59x+12?

Answer 1

Given function : $f(x)=5x^3-6x^2-59x+12.$

$\mathrm{Factor\:}5x^3-6x^2-59x+12$

Splitting terms to factor it out.

$5x^3-x^2-5x^2+x-60x+12$

Making it into groups

(5x^3-x^2) + (-5x^2+x) + (-60x+12)

Factoring out GCF of each group.

$x^2(5x-1)-x(5x-1)-12(5x-1)$

$=(5x-1)(x^2-x-12)$

Factoring x^2-x-12 into (x+3)(x-4)

$(5x-1)(x^2-x-12) = (5x-1)(x+3)(x-4)$

Applying zero product rule.

5x-1 =0

x+3=0 and

x-4= 0

On solving

x = 1/5 , x=-3 and x=4.

Therefore, zeros of the given function are $x=-3,\:x=4,\:x=\frac{1}{5}$.

Answer 2

Hello,

5x^3-6x²-59x+12=5x^3-x²-5x²+x-60x+12
=x²(5x-1)-x(5x-1)-12(5x-1)
=(5x-1)(x²-x-12)
=(5x-1)(x²+3x-4x-12)
=(5x-1)(x(x+3)-4(x+3))
=(5x-1)(x+3)(x-4)

zeros are  -3,4,1/5