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.