# How do you multiply (7x^2-2x+5)(x^2-x-7)?

May 10, 2017

Sum of the product of each term of first trinomial and the second trinomial.

#### Explanation:

Formatted question: $\left(7 {x}^{2} - 2 x + 5\right) \left({x}^{2} - x - 7\right)$

We can multiply the two trinomials together by finding the sum of the product of each term of the first trinomial to each term of the second trinomial (or the entire second trinomial):
$= 7 {x}^{2} \left({x}^{2} - x - 7\right) - 2 x \left({x}^{2} - x - 7\right) + 5 \left({x}^{2} - x - 7\right)$
$= 7 {x}^{4} - 7 {x}^{3} - 49 {x}^{2} - 2 {x}^{3} + 2 {x}^{2} + 14 x + 5 {x}^{2} - 5 x - 35$

We can reorder the expression to put $x$ terms of equal degree next to each other:
$= 7 {x}^{4} - 7 {x}^{3} - 2 {x}^{3} - 49 {x}^{2} + 2 {x}^{2} + 5 {x}^{2} + 14 x - 5 x - 35$

Now, we can combine like-terms:
$= 7 {x}^{4} - 9 {x}^{3} - 42 {x}^{2} + 9 x - 35$

May 10, 2017

See a solution process below:

#### Explanation:

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

$\left(\textcolor{red}{7 {x}^{2}} - \textcolor{red}{2 x} + \textcolor{red}{5}\right) \left(\textcolor{b l u e}{{x}^{2}} - \textcolor{b l u e}{x} - \textcolor{b l u e}{7}\right)$ becomes:

$\left(\textcolor{red}{7 {x}^{2}} \times \textcolor{b l u e}{{x}^{2}}\right) - \left(\textcolor{red}{7 {x}^{2}} \times \textcolor{b l u e}{x}\right) - \left(\textcolor{red}{7 {x}^{2}} \times \textcolor{b l u e}{7}\right) - \left(\textcolor{red}{2 x} \times \textcolor{b l u e}{{x}^{2}}\right) + \left(\textcolor{red}{2 x} \times \textcolor{b l u e}{x}\right) + \left(\textcolor{red}{2 x} \times \textcolor{b l u e}{7}\right) + \left(\textcolor{red}{5} \times \textcolor{b l u e}{{x}^{2}}\right) - \left(\textcolor{red}{5} \times \textcolor{b l u e}{x}\right) - \left(\textcolor{red}{5} \times \textcolor{b l u e}{7}\right)$

$7 {x}^{4} - 7 {x}^{3} - 49 {x}^{2} - 2 {x}^{3} + 2 {x}^{2} + 14 x + 5 {x}^{2} - 5 x - 35$

We can now group and combine like terms:

$7 {x}^{4} - 7 {x}^{3} - 2 {x}^{3} + 5 {x}^{2} - 49 {x}^{2} + 2 {x}^{2} + 14 x - 5 x - 35$

$7 {x}^{4} + \left(- 7 - 2\right) {x}^{3} + \left(5 - 49 + 2\right) {x}^{2} + \left(14 - 5\right) x - 35$

$7 {x}^{4} + \left(- 9\right) {x}^{3} + \left(- 42\right) {x}^{2} + 9 x - 35$

$7 {x}^{4} - 9 {x}^{3} - 42 {x}^{2} + 9 x - 35$