Question 91217: factoring a quadratic with leading coefficient greater than one
2y^2+15y+28 Found 2 solutions by stanbon, checkley71:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! factoring a quadratic with leading coefficient greater than one
2y^2+15y+28
ac=2*28=56
b=15
Think of two numbers whose product is 56 and whose sum is 15
The numbers are 7 and 8
Replace 15y by 7y+8y to get:
2y^2+7y+8y+28
Factor to get:
y(2y+7)+4(2y+7)
Factor again to get:
(2y+7)(y+4)
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Cheers,
Stan H.
You can put this solution on YOUR website! 2Y^2+15Y+28
FACTORS OF 2=2*1 & FACTORS OF 28=(28*1),(14*2),(7*4)
NOW FIND THE SUM OF THE PRODUCTS OF THESE FACTORS THAT=15.
2*28+1*1=56+1=57 NOT A SOLUTION
2*4+1*7=8+7=15 LOOKS LIKE A SOLUTION.
(2X+7)(X+4)
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