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September 24th, 2017, 01:38 PM   #1
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Factoring!

When factoring the RIGHT side of a trinomial if there are multiple correct answers then how is it decided which one is correct?

For example
y^2-4y-12

The answer could either be
(y-6)(y+2)
(y+6)(y-2)
(y-4)(y+3)
(y+4)(y-3)
(y-1)(y+12)
(y+1)(y-12)

I would assume the correct answer would be (y+4)(y-3) but that actually factors the whole not just the right.


The book states the correct answer as
(y+2)(y-6)

But how is that decided?
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September 25th, 2017, 02:37 AM   #2
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There is only one correct factorization ...

$y^2-4y-12=(y-6)(y+2)$

... all the rest will not yield the linear term $-4y$ when expanded (multiplied).
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September 25th, 2017, 01:04 PM   #3
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Quote:
Originally Posted by Kejntesh View Post
When factoring the RIGHT side of a trinomial if there are multiple correct answers then how is it decided which one is correct?

For example
y^2-4y-12

The answer could either be
(y-6)(y+2)
(y+6)(y-2)
(y-4)(y+3)
(y+4)(y-3)
(y-1)(y+12)
(y+1)(y-12)

I would assume the correct answer would be (y+4)(y-3) but that actually factors the whole not just the right.
Here's a tip- do not "assume" in mathematics. It is okay to "guess" if you then check. If you were to "guess" that (y+ 4)(y- 3) is a factorization of y^2- 4y- 12, you can check by doing the multiplication: (y+ 4)(y- 3)= y(y- 3)+ 4(y- 3)= y^2- 3y+ 4y- 12= y^2+ y- 12, NOT y^2- 4y- 12 so that is NOT correct.

Quote:
The book states the correct answer as
(y+2)(y-6)

But how is that decided?
(y+ 2)(y- 6)= y(y- 6)+ 2(y- 6)= y^2- 6y+ 2y- 12= y^2- 4y- 12.

In general, in order to factor y^2+ ay+ b into (y+ u)(y+ v) you need to find two numbers, u and v, such that uv= b and u+ v= a. Here, a= -4 and b= -12. How can you factor -12? One way is (-1)(12) but -1+ 12= 11, not -4. Another is, of course, (-12)(1) but -12+ 1= -11, not 4. Another is 3(-4) but 3- 4= -1 not -4. (-3)(4) is 12 but -3+ 4= 1, not -4. (-2)(6)= -12 but -2+ 6= 4 not -4. (2)(-6)= -12 and 2- 6= -4! AHA!
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