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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?
Wiki User
∙ 8y ago
In order to solve something, there needs to be an equation.
If I have read your equation right you have:
(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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