# What is the standard form of a polynomial (x-1)(x^2+6x+4)?

Jun 12, 2017

See a solution process below:

#### Explanation:

To put this expression in the standard form you must multiply these two terms by multiplying each individual term in the left parenthesis by each individual term in the right parenthesis.

$\left(\textcolor{red}{x} - \textcolor{red}{1}\right) \left(\textcolor{b l u e}{{x}^{2}} + \textcolor{b l u e}{6 x} + \textcolor{b l u e}{4}\right)$ becomes:

$\left(\textcolor{red}{x} \times \textcolor{b l u e}{{x}^{2}}\right) + \left(\textcolor{red}{x} \times \textcolor{b l u e}{6 x}\right) + \left(\textcolor{red}{x} \times \textcolor{b l u e}{4}\right) - \left(\textcolor{red}{1} \times \textcolor{b l u e}{{x}^{2}}\right) - \left(\textcolor{red}{1} \times \textcolor{b l u e}{6 x}\right) - \left(\textcolor{red}{1} \times \textcolor{b l u e}{4}\right)$

${x}^{3} + 6 {x}^{2} + 4 x - {x}^{2} - 6 x - 4$

We can now group and combine like terms:

${x}^{3} + 6 {x}^{2} - {x}^{2} + 4 x - 6 x - 4$

${x}^{3} + 6 {x}^{2} - 1 {x}^{2} + 4 x - 6 x - 4$

${x}^{3} + \left(6 - 1\right) {x}^{2} + \left(4 - 6\right) x - 4$

${x}^{3} + 5 {x}^{2} + \left(- 2\right) x - 4$

${x}^{3} + 5 {x}^{2} - 2 x - 4$