How do you determine the binomial factors of x^3+4x^2-x-4x3+4x2x4?

1 Answer
Dec 14, 2016

x^3+4x^2-x-4=(x-1)(x+1)(x+4)x3+4x2x4=(x1)(x+1)(x+4)

Explanation:

It is apparent from the given polynomial x^3+4x^2-x-4x3+4x2x4, that if x^2x2 is taken from first two terms, we have x+4x+4 left out and this is also left out if take out -11 from remaining two terms.

Hence x^3+4x^2-x-4x3+4x2x4

= x^2(x+4)-1(x+4)x2(x+4)1(x+4)

= (x^2-1)(x+4)(x21)(x+4)

= (x^2+x-x-1)(x+4)(x2+xx1)(x+4)

= (x(x+1)-1(x+1))(x+4)(x(x+1)1(x+1))(x+4)

= (x-1)(x+1)(x+4)(x1)(x+1)(x+4)