Math help please.

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Oct 19, 2000
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My wife is working on some algebra for her school and she can't figure this one problem out. It's been years since I've done any math in high school, so I obviously have no idea what to do in order to help her. I'm asking for help so she can figure out what she's doing wrong. It's her last problem on her homework.

The problem:

"Factor by changing to 2 binomials and simplifying:"

x^2-8x+15 over 2x^2-5x-25

Possible answers:

A. x-3 over 2x+5
B. x+3 over 2x-5
C. 3 over 2x-5
D. x-3 over x+5

Yes, I know this is most likely a very, very simple problem for anybody who comes into this thread, some of us don't get math as quickly as others. Could someone explain how to solve this? She doesn't just want an answer, she wants to understand what she's doing wrong.

Thanks!
 

rcpratt

Lifer
Jul 2, 2009
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(x-5)(x-3) over (2x+5)(x-5). Then cancel.

In the first, the numbers have to multiply to 15 and sum to -8. The second is harder because of the 2x^2, but you know both numbers have to be 5 since that's the only factor of 25 (besides 1 and 25). The negative has to be matched to the 2x so that the sum is -5x.

Bonus hint - you can eliminate C and D immediately because 3 isn't a possible factor of a first-order polynomial and D has no 2x, which would be very unlikely since the denominator would likely have one 2x factor while the numerator wouldn't have one.

Or just use the quadratic formula.
 
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DrPizza

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factor (unfoil) the numerator, and factor (unfoil) the denominator. Then, you can cancel out a pair of factors.

i.e. similar to this: 15/35 = (5*3)/(5*7) and the 5's cancel out.
Except, you'll have two binomials in the numerator & two in the denominator.
 

Dumac

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Dec 31, 2005
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(x-5)(x-3) over (2x+5)(x-5). Then cancel.

In the first, the numbers have to multiply to 15 and sum to -8. The second is harder because of the 2x^2, but you know both numbers have to be 5 since that's the only factor of 25 (besides 1 and 25). The negative has to be matched to the 2x so that the sum is -5x.

This. You factor the numerator and denominator
 
Oct 19, 2000
17,860
4
81
(x-5)(x-3) over (2x+5)(x-5). Then cancel.

In the first, the numbers have to multiply to 15 and sum to -8. The second is harder because of the 2x^2, but you know both numbers have to be 5 since that's the only factor of 25 (besides 1 and 25). The negative has to be matched to the 2x so that the sum is -5x.

Awesome. She had the first part but the second part was throwing her off. I read to her your explanation and she said it makes sense.

Thanks to all!



p.s. No pics of wife. :eek:
 

DrPizza

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Also, a technique that "gets the right answer" if multiple choice questions on a test:
Pick a number such as 7 for x.
Plug it in to the original expression.
x^2-8x+15 over 2x^2-5x-25
would be
7^2-8*7+15 over 2*7^2-5*7-25
=49-56+15 over 98-35-25
=8/38 (simplifies to 4/19)
But, just go with whatever decimal the calculator says if you don't want to bother with fractions..

Then, plug the same x-value into EACH of the multiple choices
A. x-3 over 2x+5 = 4/19
B. x+3 over 2x-5 = 10/9
C. 3 over 2x-5 = 3/9
D. x-3 over x+5 = (7-3)/(7+5)=4/12 = 1/3

Since A is the only one that matches, A is the answer. No algebra skills required. Note: there's a chance that more than one of the choices matches (by coincidence). If that happens, pick another number to plug in for x. So, while I wouldn't teach this as the "right" way to solve problems, I do teach it as a test taking technique for students.
 

rcpratt

Lifer
Jul 2, 2009
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Also, a technique that "gets the right answer" if multiple choice questions on a test:
Haha, I love when math teachers try to use multiple choice and make it too easy to just plug the answers like that.
 

Molondo

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Sep 6, 2005
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Our stats mid term is going to be MC, fucking shit. accuracy to 7 decimal places...
 
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