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 $x^4 - 4x^2 + 16$ in the form
$(x^2 + Ax + B)$ $(x^2 +Cx + D)$

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
Quote:

Originally Posted by dojo
Hi - I have an assignment with a question that is stumping me at the first hurdle.

Express $x^4 - 4x^2 + 16$ in the form

$(x^2 + Ax + B)$ $(x^2 +Cx + D)$ 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??

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

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

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

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

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

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

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