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# How do you solve a squared minus 4 over 2 plus 6a times 3 plus 9a over a squared plus 5a plus 6?

Updated: 9/17/2023

Wiki User

8y ago

In order to solve something, there needs to be an equation.

(a² - 4)/(2 + 6a) x (3 + 9a)/(a² + 5a + 6)

This looks like it needs to be simplified.

There are two fractions being multiplied, so multiply the numerators and denominators together, giving:

(a² - 4)/(2 + 6a) x (3 + 9a)/(a² + 5a + 6) = ((a² - 4)(3 + 9a)) / ((2 + 6a)(a² + 5a + 6))

Now, simplify the polynomials to remove the squared terms and any common factors:

a² - 4 = (a - 2)(a + 2)

2 + 6a = 2(1 + 3a)

3 + 9a = 3(1 + 3a)

a² + 5a + 6 = (a + 3)(a + 2)

Giving:

((a² - 4)(3 + 9a)) / ((2 + 6a)(a² + 5a + 6)) = ((a + 2)(a - 2)3((1 + 3a))/(2(1 + 3a)(a + 3)(a + 2))

This can now be simplified by cancelling out equivalent terms, giving:

((a + 2)(a - 2)3(1 + 3a))/(2(1 + 3a)(a + 3)(a + 2)) = ((a - 2)3) / (2(a+3)) = 3(a - 2) / 2(a + 3) = (3a - 6) / (2a + 6)

Wiki User

8y ago

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Q: How do you solve a squared minus 4 over 2 plus 6a times 3 plus 9a over a squared plus 5a plus 6?
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