# Polynomial problem

• Oct 7th 2010, 01:41 PM
dojo
Polynomial problem
Hi - I have an assignment with a question that is stumping me at the first hurdle.

Express $\displaystyle x^4 - 4x^2 + 16$ in the form
$\displaystyle (x^2 + Ax + B)$$\displaystyle (x^2 +Cx + D) where A,B,C & D are real constants. I just can't see how to get it in this form to complete the rest of the partial fraction question. Any ideas?? • Oct 7th 2010, 01:48 PM e^(i*pi) It's telling you how to express it? To find A, B, C and D you can equate the coefficients • Oct 7th 2010, 01:55 PM Archie Meade Quote: Originally Posted by dojo Hi - I have an assignment with a question that is stumping me at the first hurdle. Express \displaystyle x^4 - 4x^2 + 16 in the form \displaystyle (x^2 + Ax + B)$$\displaystyle (x^2 +Cx + D)$ just multiply this out and compare coefficients

where A,B,C & D are real constants.

I just can't see how to get it in this form to complete the rest of the partial fraction question. Any ideas??

$\displaystyle x^4-4x^2+16=\left(x^2+Ax+B\right)\left(x^2+Cx+D\right)$

$\displaystyle =x^4+Cx^3+Dx^2+Ax^3+ACx^2+ADx+Bx^2+BCx+BD$

$\displaystyle =x^4+x^3(A+C)+x^2(B+AC+D)+x(AD+BC)+BD$

which tells us that $\displaystyle A=-C$ and $\displaystyle AD=-BC$

since $\displaystyle x^4-4x^2+16$ has no $\displaystyle x^3$ or $\displaystyle x$ terms.

Also $\displaystyle B+AC+D=-4$ and $\displaystyle BD=16$

from which the values A, B, C, D are discovered.
• Oct 8th 2010, 04:40 AM
HallsofIvy
We can rewrite $\displaystyle x^4- 4x^2+ 16$ as $\displaystyle x^4+ 8x^2+ 16- 12x^2= (x^2+ 4)^2- 12x^2$ and treat it as a "difference of squares". Of course, it cannot be factored with integer coefficients.