How do you multiply $- 3 n^{2} (- 2 n^{2} + 3 n + 4)$ $-3n^2(-2n^2+3n+4)$ ?

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How do you multiply $- 3 n^{2} (- 2 n^{2} + 3 n + 4)$ $-3n^2(-2n^2+3n+4)$ ?

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Answer 1 · 2017-06-13T12:49:46

Use the distributive property.

The multiplied form is $6 n^{4} - 9 n^{3} - 12 n^{2}$ $6n^4 - 9n^3 - 12n^2$

Explanation:

The distributive property tells us that:

$a \cdot (b + c)$ $color(red)a*(b+c)$

$a \cdot b + a \cdot c$ $color(red)a*b+color(red)a*c$

So we can use this property to distribute the $- 3 n^{2}$ $-3n^2$ term:

$- 3 n^{2} (- 2 n^{2} + 3 n + 4)$ $color(red)(-3n^2)(-2n^2+3n+4)$

$= - 3 n^{2} \cdot (- 2 n^{2}) + - 3 n^{2} \cdot (3 n) + - 3 n^{2} \cdot 4$ $=color(red)(-3n^2) * (-2n^2) + color(red)(-3n^2) * (3n) + color(red)(-3n^2) * 4$

Now just multiply each group of terms together. Remember that two negatives make a positive.

$= 6 n^{4} - 9 n^{3} - 12 n^{2}$ $= 6n^4 - 9n^3 - 12n^2$

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How do you multiply $- 3 n^{2} (- 2 n^{2} + 3 n + 4)$ $-3n^2(-2n^2+3n+4)$ ?

1 Answer

Explanation:

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How do you multiply −3n2(−2n2+3n+4)-3n^2(-2n^2+3n+4)?

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How do you multiply $- 3 n^{2} (- 2 n^{2} + 3 n + 4)$ $-3n^2(-2n^2+3n+4)$ ?