Question 93571: Can you solve by factoring the quadratic
5y³ + 34y² = 7y
and show every step?
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! factor the quadratic 5y^3+34y^2=7y
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It's a cubic equation.
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Rearrange to get:
5y^3+34y^2-7y = 0
You have a common factor of "y".
y(5y^2+34y-7)=0
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To factor 5y^2+34y-7 use the AC Method:
Think of two integers whose product = AC = 5*-7 = -35
and whose sum is B = 34.
The numbers are 35 and -1
Rewrite the quadratic as:
5y^2+35y-y-7
Factor the 1st two and the last two terms separately to get:
5y(y+7)-(y+7)
Factor again to get:
(y+7)(5y-1)
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So, y(y+7)(5y-1)=0
You have factored the problem.
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If you want to "solve" the equation you get:
y=0 or y=-7 or y=1/5
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Cheers,
Stan H.
Answer by Edwin McCravy(20055)  (Show Source):
You can put this solution on YOUR website!
Can you solve by factoring the quadratic
5y³ + 34y² = 7y
and show every step?
5y³ + 34y² = 7y
Get 0 on the right by adding -7y to both sides
5y³ + 34y² - 7y = 0
y is a common factor so factor out y
y(5y² + 34y - 7) = 0
Write this:
y( y )( y ) = 0
Now think of a way to write 5, the first
coefficient as the product of two positive
integers. Since 5 is prime there is only
one way, namely 1×5 or 5×1. So put a 1
before the first y and a 5 before the second y.
(It would be OK to reverse these).
y(1y )(5y ) = 0
Now think of a way to write 7, the last
term, as the product of two positive
integers. Since 7 is also prime there is
only one way, namely 1×7 or 7×1. So try
putting a 1 at the end of the first
parentheses and a 7 at the end of the
second parentheses:
y(1y 1)(5y 7) = 0
Now check the OUTERS and INNERS as though
you were going to multiply that out by FOIL,
ignoring the signs. The OUTER product would
be 1y times 7 or 7y and the INNER product
would be 1 times 5y or 5y.
But there is no way that 7y and 5y could
combine to give the mddle term +34y no matter
what signs we gave them.
So let's discard this
y(1y 1)(5y 7) = 0
and reverse the 1 and the 7.
y(1y 7)(5y 1) = 0
Now again check the OUTERS and INNERS as though
you were going to multiply that out by FOIL,
ignoring the signs. The OUTER product would
be 1y times 1 or 1y and the INNER product
would be 7 times 5y or 35y.
This time there IS A way that 1y and 35y could
combine to give the middle term +34y. That would
be if we gave the 35y a positive sign and the 1y a
negative sign, since +35y-1y does give the middle
term +34y. So we put a + by the 7 to make the INNERS
be +35y and we put a - by the 1 in the second
parentheses to make the OUTERS be -1y
y(1y + 7)(5y - 1) = 0
Of course we can erase the 1 before the y in
the first parentheses:
y(y + 7)(5y - 1) = 0
So the left side is now completely factored.
The factors are y, y + 7, and 5y - 1
We set each factor = 0
Setting the factor y = 0 just gives y = 0
Setting the factor y + 7 = 0, gives y = -7
Setting 5y - 1 = 0 gives 5y = 1 or y = 1/5
So there are three solutions: 0, -7, and 1/5
Edwin
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