Yet another factoring problem

1. Feb 18, 2006

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.

2. Feb 18, 2006

arildno

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

3. Feb 18, 2006

Hollysmoke

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

4. Feb 18, 2006

arildno

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).

5. Feb 18, 2006

Hollysmoke

would it be 1,8 and 12?

6. Feb 18, 2006

arildno

That is correct.

7. Feb 18, 2006

Hollysmoke

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

8. Feb 18, 2006

arildno

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

9. Feb 18, 2006

Hollysmoke

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

10. Feb 18, 2006

arildno

No. Try again.

11. Feb 18, 2006

Hollysmoke

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

12. Feb 18, 2006

Hollysmoke

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

13. Feb 18, 2006

arildno

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

14. Feb 18, 2006

Hollysmoke

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

15. Feb 18, 2006

arildno

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?

16. Feb 18, 2006

Hollysmoke

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

17. Feb 18, 2006

arildno

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?

18. Feb 18, 2006

Hollysmoke

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

19. Feb 18, 2006

arildno

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

20. Feb 18, 2006

Hollysmoke

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