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Question 328707: Could I please get help with this problem. I have tried different ways to solve it and can not get it to come out. In the book it shows to use -1 in front. -1(30w²+w+1. Don't really understand.Do I need to change signs before factoring? Factor completely and check. Thank you for your time.
-30w² + w + 1
Found 2 solutions by rapaljer, Theo:
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! -30w^2+w+1
-1(30w^2-w-1)
Now, this is a factorable trinomial. You need to find two numbers whose product is 30w^2 (probably 6w*5w) like this:
-1(6w____)(5w_____)
The last times last must come out to -1, so it must be either -1*1 or 1*-1. The combination that works to give you a -w for the middle term (by trial and error!) is
-1(6w+1)(5w-1)
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Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! The equation is -30w^2 + w + 1
The reason they tell you to use a -1 in front of it is because it's easier to factor when the coefficient of the x^2 term is positive.
This can be done in the following manner.
Set your equation = to 0 to get:
-30w^2 + w + 1 = 0
Multiply both sides of this equation by -1 to get:
30w^2 - w - 1 = 0
The right sides of the equation remains the same because 0 * anything is still 0.
The left side of the equation now has a positive coefficient for the x^2 term.
You still have a problem with this equation because the coefficient of the x^2 term is not equal to 1.
Making the coefficient of the x^2 = 1 sometimes helps in factoring and is necessary if solving using the completing the squares method.
If you factor out the 30 from the left hand side of this equation, you get:
30 * (x^2 - (1/30)*x - (1/30)) = 0
If you can factor (x^2 - (1/30)*x - (1/30)) then you should be home free.
You can factor this using the quadratic equation or by completing the squares if you can't eyeball the solution.
Since 1/30 is the same as (1/5) * (1/6), it's possible that our solution will be a combination of these.
Your factors are (x - (1/5)) * (x + (1/6)) = 0.
I'm cheating now because I already know the solution. I was not able to eyeball the solution without first using the quadratic formula to solve for x.
Regardless, if these factors are good, then you should be able to substitute in the original equation to confirm that the equation is true with them.
Since the factors are (x - (1/5)) * (x + (1/6)) = 0, then:
x = (1/5) or x = (-1/6)
Substitute (1/5) in the original equation first.
The original equation is:
-30x^2 + x + 1 = 0
When x = (1/5), this equation becomes:
-30 * (1/25) + (1/5) + 1 = 0 which becomes:
-30/25 + 1/5 + 1 = 0 which becomes:
-30/25 + 5/25 + 25/25 = 0 which becomes:
-25/25 + 25/25 = 0 which is true confirming the value of x = (1/5) is good.
When x = - (1/6), this equation becomes:
-30/36 - 1/6 + 1 = 0 which becomes:
-30/36 - 6/36 + 36/36 = 0 which becomes:
-36/36 + 36/36 = 0 which is true confirming the value of x = - (1/6) is also good.
Those are your solutions.
Your original expression was:
-30x^2 + x + 1
You set it equal to 0 to get:
-30x^2 + x + 1 = 0
you factored out the (-30) to get:
(-30) * (x^2 - x/30 - 1/30) = 0
You factored x^2 - x/30 - 1/30 to get:
(-30) * (x - (1/5)) * (x + (1/6)) = 0
Those are your factors.
If you multiply these factors together, you should get your original equation.
-30 * (x - 1/5) = -30x + 6
(-30x + 6) * (x + 1/6) = -30x^2 -5x + 6x + 1 which becomes:
-30x^2 + x + 1 which is where you started from.
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