How do you simplify $- \frac{2}{3} n^{2} (- 9 n^{2} + 3 n + 6)$ $-2/3n^2(-9n^2+3n+6)$ ?

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How do you simplify $- \frac{2}{3} n^{2} (- 9 n^{2} + 3 n + 6)$ $-2/3n^2(-9n^2+3n+6)$ ?

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Answer 1 · 2017-08-19T14:47:52

$6 n^{4} - 2 n^{3} - 4 n^{2}$ $6n^4-2n^3-4n^2$

Once you are used to these you will not need to break it down into steps but do it all at once.

Explanation:

Taking it one step at a time

Think of this as:

$2 \times \frac{1}{3} \times n^{2} \times (- 1) \times (- 9 n^{2} + 3 n + 6)$ $2xx1/3xxn^2xx(-1)xx(-9n^2+3n+6)$

$Step 1$ $color(blue)("Step 1")$
Multiply everything inside the brackets by $(- 1)$ $(-1)$ giving:

$2 \times \frac{1}{3} \times n^{2} \times (+ 9 n^{2} - 3 n - 6)$ $2xx1/3xxn^2xx(+9n^2-3n-6)$

$.............................................................................$ $"............................................................................."$
$Step 2$ $color(blue)("Step 2")$
Multiply everything inside the brackets by $n^{2}$ $n^2$ giving:

$2 \times \frac{1}{3} \times (9 n^{4} - 3 n^{3} - 6 n^{2})$ $2xx1/3xx(9n^4-3n^3-6n^2)$

$.............................................................................$ $"............................................................................."$
$Step 3$ $color(blue)("Step 3")$
Multiply everything inside the brackets by $\frac{1}{3}$ $1/3$ giving:

$2 \times (3 n^{4} - 1 n^{3} - 2 n^{2})$ $2xx(3n^4-1n^3-2n^2)$

$.............................................................................$ $"............................................................................."$
$Step 4$ $color(blue)("Step 4")$
Multiply everything inside the brackets by $2$ $2$ giving:

$6 n^{4} - 2 n^{3} - 4 n^{2}$ $6n^4-2n^3-4n^2$

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How do you simplify $- \frac{2}{3} n^{2} (- 9 n^{2} + 3 n + 6)$ $-2/3n^2(-9n^2+3n+6)$ ?

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How do you simplify $- \frac{2}{3} n^{2} (- 9 n^{2} + 3 n + 6)$ $-2/3n^2(-9n^2+3n+6)$ ?