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

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Answer 1 · 2016-01-24T16:03:47

#x^3+3x^2-4x-12 = 0#

you need to find numbers such as :

#x_1 + x_2 + x_3 = -3#

#x_1x_2x_3 = 12#

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

Sum of all root = #-b#

Product of all root = #(-1)^nk#

For this take all the divisor of #+-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)