Asked by Amanda
                Factor Completly:
3y^2+7y-20
I got (5y+12)(y-1)
            
        3y^2+7y-20
I got (5y+12)(y-1)
Answers
                    Answered by
            Michael
            
    You can check your answer by foiling.
(5y + 12)(y - 1)
5y^2 - 5y + 12y - 12
5y^2 + 7y - 12
That's not our original, so it hasn't been correctly factored.
3y^2 + 7y - 20
We know the first part of each bracket because there's only one option...
(3y )(y )
Now we have to imagine foiling as we figure this out. The last two numbers multiplied together must equal -20, but 3y*(first number) + y*(second number) has to be 7y.
For -20, we can have: 5 & 4, 10 & 2, or 20 & 1 (with one of the two numbers being negative).
As a general rule, usually choose the two numbers that are closest together. In this case, that is 5 & 4.
(3y 5)(y 4)
We just have to pick our signs now. We know one must be negative, and one must be positive (because multiplied together, they must equal -20).
(3y - 5)(y + 4)
And the check by foiling...
3y^2 + 12y - 5y - 20
3y^2 + 7y - 20
That's our original, so we got it right. It cannot be factored any further.
    
(5y + 12)(y - 1)
5y^2 - 5y + 12y - 12
5y^2 + 7y - 12
That's not our original, so it hasn't been correctly factored.
3y^2 + 7y - 20
We know the first part of each bracket because there's only one option...
(3y )(y )
Now we have to imagine foiling as we figure this out. The last two numbers multiplied together must equal -20, but 3y*(first number) + y*(second number) has to be 7y.
For -20, we can have: 5 & 4, 10 & 2, or 20 & 1 (with one of the two numbers being negative).
As a general rule, usually choose the two numbers that are closest together. In this case, that is 5 & 4.
(3y 5)(y 4)
We just have to pick our signs now. We know one must be negative, and one must be positive (because multiplied together, they must equal -20).
(3y - 5)(y + 4)
And the check by foiling...
3y^2 + 12y - 5y - 20
3y^2 + 7y - 20
That's our original, so we got it right. It cannot be factored any further.
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