Answer:
4(3n+2) or 12n+8
Step-by-step explanation:
Given expression is:

The numerator of the fraction will be multiplied with 9n^2- 4
So, Multiplication will give us:

We can simplify the expression before multiplication.
The numerator will be broken down using the formula:
![a^2 - b^2 = (a+b)(a-b)\\So,\\= \frac{8[(3n)^2 - (2)^2]}{6n-4}\\ = \frac{8(3n-2)(3n+2)}{6n-4}](https://tex.z-dn.net/?f=a%5E2%20-%20b%5E2%20%3D%20%28a%2Bb%29%28a-b%29%5C%5CSo%2C%5C%5C%3D%20%5Cfrac%7B8%5B%283n%29%5E2%20-%20%282%29%5E2%5D%7D%7B6n-4%7D%5C%5C%20%3D%20%5Cfrac%7B8%283n-2%29%283n%2B2%29%7D%7B6n-4%7D)
We can take 2 as common factor from denominator

Hence the product is 4(3n+2) or 12n+8 ..
Using product rule;
f(x)=(1+6x²)(x-x²)
f'(x)=(12x)(x-x²) + (1-2x)(1+6x²) = 12x² -12x³ +1 +6x² -2x -12x³ = -24x³ +18x² -2x +1
Solving the bracket first;
f(x)=(1+6x²)(x-x²) = x -x² +6x³ -6x^4
f'(x)= 1 -2x +18x² -24x³ = -24x³ +18x² -2x +1
Answer:
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Step-by-step explanation:
<em>t would be 8 if you are multiplying .</em>