Question 110316: Factoring a quadratic wit leading coefficient greaer than 1 Found 3 solutions by wgunther, solver91311, jayanthihemkos@yahoo.com:Answer by wgunther(43) (Show Source):
You can put this solution on YOUR website! For factoring quadratics with the leading coefficent greater than 1 is easy. I'll do a harder one that the one you listed (because leading coefficent prime it's actually the same as if it was 1, it will be in the for (3w+something)(w+something) if it's factorable)
Let . Here your answer must in the form (2x+something)(2x+something) or (4x+something)(x+something) because the only way to multiple two integers to get 4 is 4 times 1 or 2 times 2. Then it's a guess and check time game. Clearly, the two contants in the factors have to be 3 and 1. So the only possibilities are (4x+1)(x+3) and (4x+3)(x+1) and (2x+1)(2x+3). Which one works?
You can put this solution on YOUR website!
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We need to find two factors of the form and . Both of the signs are + because the signs are all + in the given trinomial. We need to choose a, b, c, and d with the following relationships:
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Possibilities for a and c are 1 and 3.
Possibilities for b and d are 1 and 8 or 2 and 4.
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let's try a = 3 and c = 1. 3 * 4 = 12 and 1 * 2 = 2, 12 + 2 = 14.
So, a = 3, b = 2, c = 1, and d = 4.
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Check:
You can put this solution on YOUR website! Factoring a quadratic wit leading coefficient greaer than 1
3w^2+14w+8 =0
3w^2 +12w+2w+8 =0(factorizing the middle term)
3w(w+4)+2(w+4)=0 (taking common factor)
(3w+2)(w+4) =0
3w+2 =0
3w = 0-2
w = -2/3
w+4=0
w = 0-4
w = -4
the two roots of w are -2/3 and -4
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