Find each products.

-2/3n^2(-9n^2+3n+6)
Thanks

Is 2/3 a multiplying constant or is 3n^2 the denominator of the first fraction?

Use parentheses to clarify meaning when typing algebraic expressions.

Sorry, it actually -2/3(n^2)

(-2/3)n^2(-9n^2 + 3n +6)

= 6n^4 - 2n^3 - 4n^2

To find the product of -2/3n^2 and (-9n^2 + 3n + 6), you'll need to distribute -2/3n^2 to each term inside the parentheses.

First, let's distribute -2/3n^2 to -9n^2:

(-2/3n^2) * (-9n^2)

To multiply these terms, you can multiply the coefficients (-2/3 and -9) and add the exponents of the variable n^2.

(-2/3) * (-9) = 18/3 = 6
n^2 * n^2 = n^(2 + 2) = n^4

The first part of the product is 6n^4.

Now, let's distribute -2/3n^2 to 3n:

(-2/3n^2) * (3n)

Multiply the coefficients:

(-2/3) * 3 = -6/3 = -2

Multiply the exponents of n:

n^2 * n = n^(2 + 1) = n^3

The second part of the product is -2n^3.

Finally, distribute -2/3n^2 to 6:

(-2/3n^2) * 6

Multiply the coefficients:

(-2/3) * 6 = -12/3 = -4

The last part of the product is -4.

Now, put all the terms together:

6n^4 - 2n^3 - 4

Therefore, the product of -2/3n^2 and (-9n^2 + 3n + 6) is 6n^4 - 2n^3 - 4.