1. ## 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??

2. It's telling you how to express it?

To find A, B, C and D you can equate the coefficients

3. 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.

4. 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.