How do you factor #x^3 - 4x^2 -7x = -10#?

1 Answer
Oct 10, 2015

Answer:

#(x-1)(x-5)(x+2)=0#

Explanation:

Re-writing #x^3-4x^2-7x=-10# as
#x^3-4x^2-7x+10=0#

We notice that the sum of the coefficients is zero
#rArr x=1# is a solution to this equation
#rArr (x-1)# is a factor of #x^3-4x-7x+10#

Dividing #x^3-4x-7x+10# by #(x-1)# by polynomial long division or synthetic division gives:
#color(white)("XXX")(x-1)(x^2-3x-10)#

If there is a factoring of #(x^2-3-10)# with integer coefficients
then we need to find a pair of factors of #(-10)# which add up to #(-3)#
With a bit of thought we can come up with #(-5,2)#
which gives us the complete factoring:
#color(white)("XXX")(x-1)((x-5)(x+2)#