Question

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

Answer 1

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 + 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 + 6x`

`y = 2x^2 - 3x + 8x - 12 - 3x^2 + 6x`

We can now group and combine like terms:

`y = 2x^2 - 3x^2 - 3x + 8x + 6x - 12`

`y = (2 - 3)x^2 + (-3 + 8 + 6)x - 12`

`y = -1x^2 + 11x - 12`

`y = -x^2 + 11x - 12`