Michael W. answered • 06/23/15
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Marietta,
I thiiiink I know what they're after here.
What they've given you is a quadratic, with an x2 term, an x term, and a constant. There's a bunch of messy stuff with the a's in them, but those are the coefficients in the quadratic.
[a2 - a - 12]x2 + [2a2 - 4]x + [a2 + a]
The bolded stuff in the brackets are the coefficients. Something times x2, plus something times x, plus a constant. Still a quadratic.
The mess in front of the x2 does factor into (a-4) and (a+3). And the constant term does factor into (a) and (a+1)...so it looks like this:
[(a-4)(a+3)]x2 + [2a2 - 4]x + [(a)(a+1)]
I didn't touch the one in the middle, because when I wrote this, it got me thinking. What if I could get my head around an example without all of the variables in it? How would I factor a quadratic that had numbers in it, but with the same kind of pattern?
Here's what I came up with:
3x2 + 16x + 5
The 3 in front of the x2 factors into 3 and 1...just like in the problem, the coefficient of the x2 factors into two pieces. The constant term, 5, factors into 5 and 1...just like in the problem where we have a constant term that has two factors in it.
[3*1]x2 + 16x + [5*1]
Now, I know that you wouldn't normally write it this way, but this puppy factors, using the 3 and the 1, and then the 5 and the 1 from the constant term. You'd start with 3x and x to make 3x2...
(3x + ?) (x + ?)
And then the 5 and the 1 go where the question marks are. The question is, which way? If you put them in the right place, you can get 16x in the middle:
(3x + 1)(x + 5)
Now, take a look back at the quadratic in the problem:
[(a-4)(a+3)]x2 + [2a2 - 4]x + [(a)(a+1)]
If we use the same strategy as we did with the 3x2, we can start by looking at the x2 term. The coefficient has two factors in it, so let's split those up:
( (a-4)x + ? ) ( (a+3)x + ? )
"a-4" and "a+3" are two factors of the coefficient in front of the x2, so if we multiply them back together, we get the first term in the original quadratic.
The constant term splits up, too. (a) and (a+1). The question is, which one goes where? When we FOIL it all back out, we need to get the middle term we're looking for. Well, there are only two ways to try it, and I took a bit of an educated guess and put the "a+1" in the factor on the right. That means that the "a" goes into the factor on the left:
( (a-4)x + (a)) ( (a+3)x + (a+1))
Let's check that middle term and see if it works. When you do the "outer" and the "inner" pieces of FOIL, you get:
[(a-4)(a+1) + a(a+3)]x
Everything is multiplied by x, but the important part is simplifying the coefficient. Distribute all of the a's that are multiplied together:
[a2 - 3a - 4 + a2 + 3a]x
Combine like terms...the -3a and the 3a cancel, leaving:
[2a2 - 4]x
Ta dah! That's the middle term we were looking for.
So, final answer:
( (a-4)x + (a)) ( (a+3)x + (a+1))
I hope this helps a bit!
Michael W.
Marietta, me too. I can't say I had ever seen a problem like this before, so hopefully, laying out my thought process will help you apply the same kind of strategy when you run into foreign-looking problems. Sometimes, you have to play with it and ask yourself "have I seen something like this, maybe not exactly, but with the same kind of form/pattern?"
Good luck from the 'burbs!
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06/24/15
Marietta H.
06/23/15