What is the standard form of y=(x+4)(2x-3) -3x^2+6xy=(x+4)(2x3)3x2+6x?

1 Answer
Apr 29, 2017

See the entire solution process below:

Explanation:

First, multiply the two terms in parenthesis. To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

y = (color(red)(x) + color(red)(4))(color(blue)(2x) - color(blue)(3)) - 3x^2 + 6xy=(x+4)(2x3)3x2+6x becomes:

y = (color(red)(x) xx color(blue)(2x)) - (color(red)(x) xx color(blue)(3)) + (color(red)(4) xx color(blue)(2x)) - (color(red)(4) xx color(blue)(3)) - 3x^2 + 6xy=(x×2x)(x×3)+(4×2x)(4×3)3x2+6x

y = 2x^2 - 3x + 8x - 12 - 3x^2 + 6xy=2x23x+8x123x2+6x

We can now group and combine like terms:

y = 2x^2 - 3x^2 - 3x + 8x + 6x - 12y=2x23x23x+8x+6x12

y = (2 - 3)x^2 + (-3 + 8 + 6)x - 12y=(23)x2+(3+8+6)x12

y = -1x^2 + 11x - 12y=1x2+11x12

y = -x^2 + 11x - 12y=x2+11x12