N/A Factoring N/A

Terry

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Terry
Help please.

-9m^2(2m^2+6m-1)

I need to factor this.

Thanks,
Terry
 
Homework, eh?

What are the factors for the highest and lowest terms?

There are only so many ways to put them together.
 
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Well the factoring should look something like this:

(2m-1) (3m+1) but I am having trouble getting the 6m in the middle.
 
Homework is not for me. For 33 year old daughter and I just grabbed the 1st problem.

Has been 50 years since I did this.

Anyway, help please.
 
Quadratic formula!

Given: ax^2 + bx + c

x = (-b +/- sqrt(b^2 - 4ac))/2a

So, for your formula: 9m^2 (2m^2 +6m -1)

you already have factors of 9m and m, the problem is to factor

2m^2 +6m -1

Using the quadratic formula:
a=2, b=6, c=-1

x = (-6 +/- sqrt (6^2 - 4(2)(-1)))/(2*2)

x = -6/4 +- sqrt (44)/4

x = -3/2 +- 1/2sqrt(11)

writing as factors:

(x+3/2+sqrt(11)/2)(x+3/2-sqrt(11)/2)

multiplying that out gives (2x^2 + 6x -1)/2

Final answer in factored form:

(18)(m)(m)(m+3/2+sqrt(11)/2)(m+3/2-sqrt(11)/2)

QED!

BTW, the roots or zero crossings are:

m=0
m=-3/2 + sqrt(11)/2
m=-3/2 - sqrt(11)/2

The sqrt(11) means that it would have been very difficult to do the simple factoring into (x+something)(x+somethingelse). I'm surprised that this was given as a factoring problem because the quadratic formula was required to solve it. Is it possible that there is an error in how the original problem was entered?
 
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:blush:Jim,

No, the problem is correct. The daughter and I was looking through her new math book. She is working on her degree and I just grabbed the problem from the book and tried to factor it first.

When I went to the "Problem Solving" site and entered just the (2m^2+6m-1) , it dawned on me it was a Quadratic.

I then solved the problem just like you did above but figured no one was interested, so I didn't say anything.

I felt sort of embarrassed for posting without thinking it through. :blush:
Anyway, thanks for the help.

Terry
 
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