What is the standard form of y= (4x-4)(x^2+5x-5)?

1 Answer
Dec 27, 2015

y=4x^3+16x^2-40x+20

Explanation:

For convenience, separate out the scalar factor 4 temporarily while multiplying out, group the terms in descending degree and combine. For illustration I have shown more steps than normal:

(4x-4)(x^2+5x-5)

=4(x-1)(x^2+5x-5)

=4(x(x^2+5x-5)-1(x^2+5x-5))

=4((x^3+5x^2-5x)-(x^2+5x-5))

=4(x^3+5x^2-5x-x^2-5x+5)

=4(x^3+(5x^2-x^2)+(-5x-5x)+5)

=4(x^3+(5-1)x^2+(-5-5)x+5)

=4(x^3+4x^2-10x+5)

=4x^3+16x^2-40x+20

Alternatively, just look at the combinations of terms to give each power of x in descending order like this:

(4x-4)(x^2+5x-5)

=4x^3+(20-4)x^2-(20+20)x+20

=4x^3+16x^2-40x+20