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Earn +20 ptsQ: X plus 3 x-2 distribute
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How do you simplify x2-9 over x2-6x plus 9?
First you distribute: x2-9 = (x-3)(x+3) and that is you numerator. Distribute x2-6x+9. Find two #'s that mult. to make 9 and add to make 6. 3x3=9 and 3+3=6. now you replace the 6x with -3x and -3x so rewrite the problem as x2-3x-3x+9 then you distribute x(x-3)-3(x-3) so x2-6x+9 becomes (x-3)(x-3) and that is your denominator. Tharefore: (x-3)(x+3) over (x-3)(x-3) then cross out like terms and your answer is (x+3) over (x-3)
What is x2 plus 3x plus x plus 3?
X2+3x+x+3=x(x+1)+3(x+1)=(x+3)(x+1)
X2 plus 10x plus 21 by x plus 3?
(x2+10x+21)/(x+3) = (x+7)
What is x2 plus 2x plus 15 divided by x-3?
(x2 + 2x + 15)/(x - 3) = x + 5 + 30/(x - 3) where x ≠3
X2 plus 4x - 9 equals 5x plus 3?
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
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