• Oct 4th 2008, 05:03 AM
oxrigby
Hi I cant seem to solve this equation. Wikipedia said to sub $\displaystyle z=x^2$ but I don't know how to then get the x results back. Once i have done this I get confused as to how to put it back into two quadratics which is what i've been aed to do.

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

thnx.
• Oct 4th 2008, 05:13 AM
Simplicity
Quote:

Originally Posted by oxrigby
Hi I cant seem to solve this equation. Wikipedia said to sub $\displaystyle z=x^2$ but I don't know how to then get the x results back. Once i have done this I get confused as to how to put it back into two quadratics which is what i've been aed to do.

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

thnx.

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

Comparing $\displaystyle x^3$ coefficient: $\displaystyle 0 = C + A$
Comparing $\displaystyle x^2$ coefficient: $\displaystyle -4 = D+AC+B$
Comparing $\displaystyle x$ coefficient: $\displaystyle 0 = AD + BC$
Comparing constant coefficient: $\displaystyle 16 = BD$

Solve for $\displaystyle A,B,C, D$ and substitute this value into: $\displaystyle (x^2+Ax+B)(x^2+Cx+D)$
• Oct 4th 2008, 05:26 AM
kbartlett
If u sub z into $\displaystyle x^4-4x^2+16$ u get $\displaystyle z^2-4z+16$ because $\displaystyle z^2=x^4$.
$\displaystyle z^2-4z+16$ factorises into $\displaystyle (z-2-i\sqrt12)(z-2+i\sqrt12)$

So
factorises into $\displaystyle (x^2-2-i\sqrt12)(x^2-2+i\sqrt12)$

So A=C=0
B=$\displaystyle -2-i\sqrt12$
and D=$\displaystyle -2+i\sqrt12$
• Oct 4th 2008, 05:51 AM
oxrigby
Question says,'where ABCD are real constants'''.??
• Oct 4th 2008, 06:23 AM
kbartlett
Quote:

Originally Posted by oxrigby
Question says,'where ABCD are real constants'''.??

Then
A=$\displaystyle \sqrt12$
B=4
C=$\displaystyle -\sqrt12$
D=4

$\displaystyle (x^2-2-i\sqrt12)(x^2-2+i\sqrt12)$ = $\displaystyle (x^2+4x-\sqrt12)(x^2+4x+\sqrt12)$ = $\displaystyle x^4-4x^2+16$
• Oct 4th 2008, 06:54 AM
Opalg
Quote:

Originally Posted by oxrigby
Express $\displaystyle x^4-4x^2+16$ in the form $\displaystyle (x^2+Ax+B)(x^2+Cx+D)$

Difference of two squares:

\displaystyle \begin{aligned}x^4-4x^2+16 &= (x^2+4)^2 - 12x^2 \\ &= (x^2+4+\sqrt{12}x)(x^2+4-\sqrt{12}x) \\ &= (x^2+2\sqrt3x+4)(x^2-2\sqrt3x+4).\end{aligned}