• Feb 1st 2011, 06:35 AM
JIBTEXHNKA
Ok, this is super basic... I've been out of school for 5 years and decided to go back. I'm taking an online placement test to guesstimate where I should be and I'm stumped on one of the questions (a few, actually, but this one is really bugging me).

The problem:
(Solving polynomial equations by factoring)
7. The solutions of x2 - 2x – 24 = 24 are:
A. 6,-4
B. 8, -6
C. 2, 48
D. 20, 30
E. 0,2

I worked that out to
x2 - 2x – 48
and then since

-8 x 6 = -48
and
-8 + 6 = -2
I figure the factors are
(x-8)(x+6)
-8, 6

...right?

BUT that's not a given choice... only 8, -6. Am I missing something simple that switches the signs, or should did the test writer screw up? I have absolutely no idea where I went wrong... it seems pretty simple.

Help?
• Feb 1st 2011, 06:42 AM
Quote:

Originally Posted by JIBTEXHNKA
Ok, this is super basic... I've been out of school for 5 years and decided to go back. I'm taking an online placement test to guesstimate where I should be and I'm stumped on one of the questions (a few, actually, but this one is really bugging me).

The problem:
(Solving polynomial equations by factoring)
7. The solutions of x2 - 2x – 24 = 24 are:
A. 6,-4
B. 8, -6
C. 2, 48
D. 20, 30
E. 0,2

I worked that out to
x2 - 2x – 48
and then since

-8 x 6 = -48
and
-8 + 6 = -2
I figure the factors are
(x-8)(x+6)
-8, 6

...right? no!

BUT that's not a given choice... only 8, -6. Am I missing something simple that switches the signs, or should did the test writer screw up? I have absolutely no idea where I went wrong... it seems pretty simple.

Help?

You have

$x^2-2x-48=0\Rightarrow\ (x-8)(x+6)=0$

and

8-8=0

6-6=0

so the solutions are the values of x that make either factor zero, which are

$x=8,\;\;\;x=-6$
• Feb 1st 2011, 06:45 AM
JIBTEXHNKA
OH RIGHT. Duh. ...Did I mention it's been a while?

Thank you!