
Factor theorem!
ok, im stuck with this question:
Find the values of p and q if 4x^24x3 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 (x1)(2x3)(2x+1)^2
I got the answers from the answer key behind the book
Its due tomorrow so can someone please help meee?

I am getting 14 and 4.
$\displaystyle 8x^{4}12x^{3}6x^{2}+x+3$ has 2 real solutions and two nonreal. Are you sure you typed it correctly. One sign is wrong then it's all wrong.
If you factor $\displaystyle 4x^{2}4x3=(2x3)(2x+1)$
As you can see, two soluitons are 3/2 and 1/2
Plug these into the quartic and you get:
$\displaystyle 27p+18q+360=0$
$\displaystyle p+2q+24=0$
Solve these two for p and q and you get q=14 and p=4
Sub those in and you get:
$\displaystyle 8x^{4}4x^{3}14x^{2}+x+3=(x+1)(2x3)(2x1)(2x+1)$
The factorization you said was in the book when expanded gives:
$\displaystyle 8x^{4}12x^{3}6x^{2}+\boxed{7}x+3$

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 $\displaystyle 8x^44x^310x^2x3 $when i expanded$\displaystyle (2x^2x3)(4x^21^2) $since $\displaystyle (x+1)(2x3) = 2x^2x3$ and $\displaystyle (2x1)(2x+1)= (4x^21^2)$

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 $\displaystyle 4x^{2}4x3$. You now have $\displaystyle 2x^{2}x3$

hmmm...for that, i got $\displaystyle (4x^21^2)(2x^2x3) = 8x^44x^312x^2+2x^2x3 = 8x^44x^310x^2x3 $
and for $\displaystyle (2x^2+x1)(4x^24x3)$, i got = $\displaystyle 8x^44x^314x^2+x+3$ O_O