# How do you factor y=4x^3+12x^2-16x-48 ?

Jan 24, 2016

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

you need to find numbers such as :

${x}_{1} + {x}_{2} + {x}_{3} = - 3$

${x}_{1} {x}_{2} {x}_{3} = 12$

Because for every monic polynomial of degree $n$ you have

Sum of all root = $- b$

Product of all root = ${\left(- 1\right)}^{n} k$

For this take all the divisor of $\pm 12$

-12 ; -6 ; -4 ; -3 ; -2 ; -1 ; 1 ; 2 ; 3 ; 4 ; 6 ; 12

Sum must be negative so one root is negative, BUT, product is positive so, at least two roots are negative

you find

x_1 = -3 ; x_2 = -2 ; x_3 = 2

Of course this method work if coefficient are integers

So :

(x+3)(x+2)(x-2)