# How to find the zeros of an equation?

• Apr 4th 2010, 04:00 PM
nascar77
How to find the zeros of an equation?
How would I find these zeros? (Worried) I tried factoring by grouping but it doesnt give me the correct answer:

y=-4x^3+12x^2+96x+112

The first term is negative
• Apr 4th 2010, 04:10 PM
skeeter
Quote:

Originally Posted by nascar77
How would I find these zeros? (Worried) I tried factoring by grouping but it doesnt give me the correct answer:

y=-4x^3+12x^2+96x+112

The first term is negative

$y = -4(x^3 - 3x^2 - 24x - 28)$

using the rational root theorem, $x = -2$ is a zero for $(x^3 - 3x^2 - 24x - 28)$ ... you should be able to find the two other zeros with that piece of information.
• Apr 4th 2010, 04:15 PM
TKHunny
#1 - Simplify your life. Same solutions as this: 0 = x^3 - 3x^2 - 24x - 28

#2 - Rule of Signs: One Positive Real. Two or Zero Negative

#3 - Rational Roots: +/-1, +/-2, +/-4, +/-7, +/-14, +/-28
--------- 7 Works

#4 - Reduction: (x^3 - 3x^2 - 24x - 28)/(x - 7) = (x^2 + 4x + 4)

#5 - That is easily factored.