i have finals coming up and dont understand/forget this part of my review, please help, thanks
x / (x^2-4) + (3x - 5) / x^2 + 4x + 4
x / (x^2-4) + (3x - 5) / x^2 + 4x + 4
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Hey, I'm in Algebra 2 also :)
x / (x^2-4) + (3x-5) / (x^2+4x+4)
First you want to factor everything possible. x^2 - 4 = (x+2)(x-2) because it's a difference of two squares, and x^2 + 4x + 4 = (x+2)(x+2).
x / (x-2)(x+2) + (3x-5) / (x+2)(x+2)
Then, you have to find a common denominator. Multiply the first fraction by (x+2) and the second fraction by (x-2)
x(x+2) / (x-2)(x+2)(x+2) + (3x-5)(x-2) / (x-2)(x+2)(x+2)
Now you're ready to add them together.
x(x+2)+(3x-5)(x-2) / (x-2)(x+2)(x+2)
Multiply everything out.
x^2 + 2x + 3x^2 - 11x + 10 / x^3 + 2x^2 - 4x - 8
Simplify, or add like terms together.
4x^2 - 9x + 10 / x^3 + 2x^2 - 4x - 8
So that's your answer:
(4x^2 - 9x + 10) / (x^3 + 2x^2 - 4x - 8)
EDIT: If you want to leave it factored, you could just say:
(4x^2 - 9x + 10) / (x - 2)(x + 2)(x + 2)....which is (4x^2 - 9x + 10) / (x - 2)(x + 2)^2
But I'm sure the version that is not factored also works, as both are correct.
x / (x^2-4) + (3x-5) / (x^2+4x+4)
First you want to factor everything possible. x^2 - 4 = (x+2)(x-2) because it's a difference of two squares, and x^2 + 4x + 4 = (x+2)(x+2).
x / (x-2)(x+2) + (3x-5) / (x+2)(x+2)
Then, you have to find a common denominator. Multiply the first fraction by (x+2) and the second fraction by (x-2)
x(x+2) / (x-2)(x+2)(x+2) + (3x-5)(x-2) / (x-2)(x+2)(x+2)
Now you're ready to add them together.
x(x+2)+(3x-5)(x-2) / (x-2)(x+2)(x+2)
Multiply everything out.
x^2 + 2x + 3x^2 - 11x + 10 / x^3 + 2x^2 - 4x - 8
Simplify, or add like terms together.
4x^2 - 9x + 10 / x^3 + 2x^2 - 4x - 8
So that's your answer:
(4x^2 - 9x + 10) / (x^3 + 2x^2 - 4x - 8)
EDIT: If you want to leave it factored, you could just say:
(4x^2 - 9x + 10) / (x - 2)(x + 2)(x + 2)....which is (4x^2 - 9x + 10) / (x - 2)(x + 2)^2
But I'm sure the version that is not factored also works, as both are correct.
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[x / (x^2 - 4)] + [(3x - 5) / (x^2 + 4x + 4)]
[x / (x + 2)(x - 2)] + [(3x - 5) / (x + 2)(x + 2)
[x(x + 2) + (3x - 5)(x - 2)] / [(x + 2)(x + 2)(x - 2)]
(x^2 + 2x + 3x^2 - 6x - 5x + 10) / [(x + 2)(x^2 - 4)]
(4x^2 - 9x + 10) / (x^3 - 4x + 2x^2 - 8)
(4x^2 - 9x + 10) / (x^3 + 2x^2 - 4x - 8)
[x / (x + 2)(x - 2)] + [(3x - 5) / (x + 2)(x + 2)
[x(x + 2) + (3x - 5)(x - 2)] / [(x + 2)(x + 2)(x - 2)]
(x^2 + 2x + 3x^2 - 6x - 5x + 10) / [(x + 2)(x^2 - 4)]
(4x^2 - 9x + 10) / (x^3 - 4x + 2x^2 - 8)
(4x^2 - 9x + 10) / (x^3 + 2x^2 - 4x - 8)
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x / (x^2-4) + (3x - 5) / x^2 + 4x + 4 =
x / (x-2)(x+2) + (3x - 5)/(x+2)^2 =
[x(x+2) + (3x-5)(x-2)] / (x-2)(x+2)^2 =
(x^2 + 2x + 3x^2 - 6x - 5x + 10) / (x-2)(x+2)^2 =
(4x^2 - 9x + 10) / (x - 2)(x + 2)^2
x / (x-2)(x+2) + (3x - 5)/(x+2)^2 =
[x(x+2) + (3x-5)(x-2)] / (x-2)(x+2)^2 =
(x^2 + 2x + 3x^2 - 6x - 5x + 10) / (x-2)(x+2)^2 =
(4x^2 - 9x + 10) / (x - 2)(x + 2)^2
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Hey, I'm in Algebra 2 also :)