How do you factor #x^3 + 11x^2 + 28x#?

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Answer 1 · 2015-06-17T11:24:22

#x^3+11x^2+28x = x(x^2+11x+28) = x(x+7)(x+4)#

First identify the greatest common factor of all the terms, which is #x# and separate that out as a factor:

#x^3+11x^2+28x = x(x^2+11x+28)#

The remaining quadratic factor must have factors of the form:

#(x+a)#, #(x+b)#, with #a+b = 11# and #a * b = 28#,

since #(x+a)(x+b) = x^2+(a+b)x+ab#

The pair #7, 4# works.

Hence:

#x^2+11x+28 = (x+7)(x+4)#

and

#x^3+11x^2+28x = x(x^2+11x+28) = x(x+7)(x+4)#