# Yet another factoring problem (1 Viewer)

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#### Hollysmoke

(a^2-5a)^2 + 8(a^2-5a)+12

Since it's a trinomial, I tried to do decomposition but it didn't work. I tried to factor out the (a^2-5a) but I end up with a different answer then on the sheet.

#### arildno

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Set y=a^2-5a.
Factorize your expresion in terms of y first

#### Hollysmoke

so would it be ax^2+bx+c=0?

#### arildno

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x presumably meaning y, yes, you'll see how to factorize the polynomial once you've found the roots of your cited equation, with the relevant values for "a,b,c"
("a" NOT meaning the same here as above).

#### Hollysmoke

would it be 1,8 and 12?

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That is correct.

#### Hollysmoke

But where do I go from there? My friendsa nd I are stumped. Our teacher never taught us this >_>

#### arildno

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What are the roots of the equation:
$$y^{2}+8*y+12=0$$?
How can you therefore factorize the polynomial $y^{2}+8y+12$?

(y+4)(y+3)=0

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No. Try again.

err whoops
(y+2)(y+6)=0

#### Hollysmoke

I got the decomposition mixed up (which term to add up and which to multiply)

#### arildno

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So, you've factorized the left hand side of your EQUATION correctly.
How is thereby the POLYNOMIAL $y^{2}+8y+12$ factorized?

#### Hollysmoke

I'm not sure what you are asking for, I thought I just factored the polynomial

#### arildno

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OK:
For ANY number y, we have the identity $y^{2}+8y+12=(y+2)(y+6)[/tex] This is to factorize the POLYNOMIAL. The two numbers -2 and -6 are the solutions Y for the equation [itex]Y^{2}+8Y+12=0$

Do you agree with that?

#### Hollysmoke

Right, I understand that part. (Sorry if I'm a bit slow)

#### arildno

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That's okay.
Now that you've factorized your polynomial in y, substitute into its factorized form (y+2)(y+6) on the y-places y=a^2-5a.
What do you get then?
In particular, can you do some further factorizations?

#### Hollysmoke

so would it be (a^2-5a)(a+2)(a+6)?

#### arildno

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Eeh, why??
Have you placed the expression a^2-5a into the y-places correctly, do you think?

#### Hollysmoke

OH....is it...(a^2-5a)(a-2)(a-3)?

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