How do you multiply $2 m^{2} (2 m^{2} + 3 m - 5)$ $2m^2(2m^2+3m-5)$ ?

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Answer 1 · 2016-10-11T20:06:54

$4 m^{4} + 6 m^{3} - 10^{2}$ $4m^4+6m^3-10^2$

Distribute the term outside of the set of parenthesis to each term on the inside of the parenthesis.

So, start with the first term in the parenthesis: $2 m^{2}$ $2m^2$

Distribute $2 m^{2}$ $2m^2$ to $2 m^{2}$ $2m^2$ and multiply
$2 m^{2} \times 2 m^{2} = 4 m^{4}$ $2m^2xx2m^2=4m^4$

Next distribute $2 m^{2}$ $2m^2$ to the second term in the parenthesis: $3 m$ $3m$

$2 m^{2} \times 3 m = 6 m^{3}$ $2m^2xx3m=6m^3$

Lastly, distribute $2 m^{2}$ $2m^2$ to the third and final term in the parenthesis: $- 5$ $-5$

$2 m 62 \times - 5 = - 10 m^{2}$ $2m62xx-5=-10m^2$

So, the equation would look like:
$4 m^{4} + 6 m^{3} - 10^{2}$ $4m^4+6m^3-10^2$

I just expanded it so you could see each individual step

How do you multiply 2m^2(2m^2+3m-5)2m2(2m2+3m−5)2m^2(2m^2+3m-5)?