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

Jan 21, 2018

See a solution process below:

Explanation:

We can expand the terms on the right side of the equation by multiplying each term in the parenthesis on the left by each term in the parenthesis on the right:

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

$y = \left(\textcolor{red}{3 x} \times \textcolor{b l u e}{{x}^{2}}\right) - \left(\textcolor{red}{3 x} \times \textcolor{b l u e}{2 x}\right) + \left(\textcolor{red}{3 x} \times \textcolor{b l u e}{4}\right) - \left(\textcolor{red}{4} \times \textcolor{b l u e}{{x}^{2}}\right) + \left(\textcolor{red}{4} \times \textcolor{b l u e}{2 x}\right) - \left(\textcolor{red}{4} \times \textcolor{b l u e}{4}\right)$

$y = 3 {x}^{3} - 6 {x}^{2} + 12 x - 4 {x}^{2} + 8 x - 16$

We can now group and combine like terms:

$y = 3 {x}^{3} - 6 {x}^{2} - 4 {x}^{2} + 12 x + 8 x - 16$

$y = 3 {x}^{3} + \left(- 6 - 4\right) {x}^{2} + \left(12 + 8\right) x - 16$

$y = 3 {x}^{3} + \left(- 10\right) {x}^{2} + 20 x - 16$

$y = 3 {x}^{3} - 10 {x}^{2} + 20 x - 16$