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Question 101639: A) Factor each polynomial completely, given that the binomial following it is a factor of the polynomial.
X^3 + 2x^2 – 5x – 6, x+3
B) Factor each difference or sum of cubes.
U^3 – 125v^3
C) Factor each polynomial completely. If a polynomial is prime, say so.
1) 32a^2 + 4a -6
2) 9bn^3 + 15bn^2 – 14bn
Any help would be greatly appreciated. I need to figure these out so I can finish my project. Thank you so much. Step by step explanations would be even better, so I can learn how to do this on my own.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A) Factor each polynomial completely, given that the binomial following it is a factor of the polynomial.
X^3 + 2x^2 – 5x – 6, x+3
Using synthetic division:
-3)....1....2....-5....-6
........1....-1....-2..|..0
Quotient: x^2-x-2
Remainder: 0
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Factoring the Quotient:
(x-2)(x+1)
Therefore: f(x) = (x+3)(x-2)(x+1)
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B) Factor each difference or sum of cubes.
Form: a^3-b^3 = (a-b)(a^2+ab+b^2
Your Problem:
U^3 – 125v^3 = (U-5V)(U+5VU+25v^2)
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C) Factor each polynomial completely. If a polynomial is prime, say so.
1) 32a^2 + 4a -6
= 2(16a^2+2a-3)
= 2(16a^2+8a-6a-3)
= 2(8a(2a+1)-3(2a+1)
= 2(2a+1)(8a-3)
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2) 9bn^3 + 15bn^2 – 14bn
= bn(9n^2+15n-14)
= bn(3n-2)(3n+7)
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Cheers,
Stan H.
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