# Factor theorem!

• Feb 16th 2008, 03:57 AM
teddybear67
Factor theorem!
ok, im stuck with this question:

Find the values of p and q if 4x^2-4x-3 is a factor of the expression 8x^4 + px^3 + qx^2 + x + 3. Hence, factorise the expression completely

It is based on factor theorem...if u've ever heard of it...

The answer is p= -12, q= -6 and the factors are (x-1)(2x-3)(2x+1)^2

• Feb 16th 2008, 04:26 AM
galactus
I am getting -14 and -4.

$8x^{4}-12x^{3}-6x^{2}+x+3$ has 2 real solutions and two non-real. Are you sure you typed it correctly. One sign is wrong then it's all wrong.

If you factor $4x^{2}-4x-3=(2x-3)(2x+1)$

As you can see, two soluitons are 3/2 and -1/2

Plug these into the quartic and you get:

$27p+18q+360=0$

$-p+2q+24=0$

Solve these two for p and q and you get q=-14 and p=-4

Sub those in and you get:

$8x^{4}-4x^{3}-14x^{2}+x+3=(x+1)(2x-3)(2x-1)(2x+1)$

The factorization you said was in the book when expanded gives:

$8x^{4}-12x^{3}-6x^{2}+\boxed{7}x+3$
• Feb 16th 2008, 05:01 AM
teddybear67
wow...I get you. Thanks!!! I checked the book and it said 8x^4+px^3+qx^2+x+3
so i guess they have printing error...

but i dont get how you expand it though...i got different answers...

I got $8x^4-4x^3-10x^2-x-3$when i expanded $(2x^2-x-3)(4x^2-1^2)$since $(x+1)(2x-3) = 2x^2-x-3$ and $(2x-1)(2x+1)= (4x^2-1^2)$
• Feb 16th 2008, 05:04 AM
galactus
It's not -10x^2, it's -14x^2. Be careful. One number, one sign and it's all kaput.

Also, you have negatives where plus should be. You are using different expressions now. Which is which?.

I thought it was $4x^{2}-4x-3$. You now have $2x^{2}-x-3$
• Feb 16th 2008, 05:16 AM
teddybear67
hmmm...for that, i got $(4x^2-1^2)(2x^2-x-3) = 8x^4-4x^3-12x^2+2x^2-x-3 = 8x^4-4x^3-10x^2-x-3$

and for $(2x^2+x-1)(4x^2-4x-3)$, i got = $8x^4-4x^3-14x^2+x+3$ O_O