Question 91137: Factoring a quadratic with leading coefficient greater than one-- can your calculators answer these problems, so far answers are all wrong
10w^2+11w-6
Found 2 solutions by checkley71, stanbon:
Answer by checkley71(8403)   (Show Source):
You can put this solution on YOUR website! 10W^2+11W-6
FACTORS OF 10 ARE 10&1, -10&-1, 5&2, -5&-2
FACTORS OF -6 ARE -1&6, 1&-6, -2&3, 2&-3
NOW WE NEED TO PICK A PAIR OF X + Y FACTORS THAT WNEM MULTIPLIED ADD UP TO 11
(10*-1)+(-1*-6)=-10+6=4 NOT A GOOD PICK.
(5*-2)+(2+3)=-10+6=-4 NOT A GOOOD PICK.
(5*3)+(2*-2)=15-4=11 LOOKS LIKE A GOOD PICK. THUS THE FACTORS ARE:
(5W-2)(2W+3) ANSWER.
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Factoring a quadratic with leading coefficient greater than one-- can your calculators answer these problems, so far answers are all wrong
10w^2+11w-6
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The following is called The AC-Method:
Think of two numbers whose product = ac = 10*-6 = -60
and whose sum = b = 11
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The numbers are 15 and -4
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In place of 11w in the original problem put 15w-4w, as follows:
10w^2+15w-4w-6
Factor the 1st two and the last two terms separately:
5w(2w+3)-2(2w+3)
Factor again to get the final answer:
(2w+3)(5w-2)
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Cheers,
Stan H.
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