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Need help solving an equation by factoring.

PROBLEM: $3v^2-10v-12 = -28v + 36$

This is my solution, but it seems a little too much for school:

$3v^2+18v-48 = 0$

$3(v^2+18v-48) = 0$

$3((-1v + -8)(-1v + 2)) = 0$

$-1v + -8 = 0$

$v = -8$

$-1v + 2 = 0$

$v = 2$

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I don't know if it's too much for school, but it's wrong for math. Careful when you factor out that $3$! – Gerry Myerson Jan 29 '13 at 3:19
Your second equation is incorrect. Then the factorization is mysteriously correct. – 1015 Jan 29 '13 at 3:26
up vote 2 down vote accepted

Your algebra went astray: from $3v^2-10v-12=-28v+36$ you should get


Then you can factor out the $3$ to get $3(v^2+6v-16)=0$, so $v^2+6v-16=0$, and this then factors nicely to give you $(v+8)(v-2)=0$.

You actually made compensating errors, since $3(v^2+18v-48)$ does not factor as


your $$3\big((−1v+−8)(−1v+2)\big)=0$$ is actually equivalent to


which is essentially what I got.

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OP got the right factors, so the misfactoring of the $3$ looks like a transcription error. – Gerry Myerson Jan 29 '13 at 3:21
My answer is probably wrong because I've been using online calculators to check my work. I guess I can't rely on them. – 12japerk Jan 29 '13 at 3:23
@Gerry: I realized that just after I posted the original version; I can’t tell whether it’s a transcription error or just sloppy notation $-$ I’ve seen plenty of both! – Brian M. Scott Jan 29 '13 at 3:25
Thanks for your post Brian, It helped a lot. – 12japerk Jan 29 '13 at 3:26
@12japerk: You’re welcome. – Brian M. Scott Jan 29 '13 at 3:26

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