Factoring (a²−5a)² + 8(a²−5a) + 12 with substitution

[Hollysmoke]

I asked my friends about this and they were stumped too.

(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]

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

 

[Hollysmoke]

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

 

[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).

 

[Hollysmoke]

would it be 1,8 and 12?

 

[arildno]

That is correct.

 

[Hollysmoke]

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

 

[arildno]

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

 

[Hollysmoke]

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

 

[arildno]

No. Try again.

 

[Hollysmoke]

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

 

[Hollysmoke]

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

 

[arildno]

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]

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

 

[Hollysmoke]

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

 

[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?

 

[Hollysmoke]

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

 

[arildno]

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)?

 

[arildno]

No.
You have (y+2)(y+6)
If you substitute the y's in that expression using the equality y=a^2-5a, what do you get?

 

[VietDao29]

You should read all the posts here again, Hollysmoke:
You need to factor $$(a ^ 2 - 5a) ^ 2 + 8(a ^ 2 - 5a) + 12$$.
So you notice that a² - 5a appears twice, you may want to make your expression have a nicer look.
So you'll let: $$y = a ^ 2 - 5a$$, your expression suddenly becomes:
y² + 8y + 12, which can be factored to (y + 2) (y + 6). It looks nicer, right?
Now, having let: y = a² - 5a, you should substitute that back to the expression (y + 2) (y + 6) to get some expression in terms of a, instead of y.
After substituting, can the expression can be factored more?

 

[Hollysmoke]

oops sorry...I forgot to post my Thanks >_<

 

[arildno]

Okay, do you agree that with substituting a^2-5a for y, we get:
$$(a^{2}-5a+2)(a^{2}-5a+6)$$ ?

 

[Hollysmoke]

Yup. Then u use decomposition for the 2nd term, resulting in (a-2)(a-3)

 

[arildno]

Quite so. Thus, you've got the decompositions:
$$(a^{2}-5a)^{2}+8(a^{2}-5a)+12=(a^{2}-5a+2)(a^{2}-5a+6)=(a^{2}-5a+2)(a-2)(a-3)$$
Now, try and decompose the factor [itex](a^{2}-5a+2)[/tex] in a similar manner, and you're done.[/itex]