biquadratics

**oxrigby** · October 4th 2008, 05:03 AM

Hi I cant seem to solve this equation. Wikipedia said to sub $z=x^2$ 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 $x^4-4x^2+16$ in the form $(x^2+Ax+B)(x^2+Cx+D)$

thnx.

**Simplicity** · October 4th 2008, 05:13 AM

Originally Posted by oxrigby

Hi I cant seem to solve this equation. Wikipedia said to sub $z=x^2$ 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 $x^4-4x^2+16$ in the form $(x^2+Ax+B)(x^2+Cx+D)$

thnx.

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

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

Solve for $A,B,C, D$ and substitute this value into: $(x^2+Ax+B)(x^2+Cx+D)$

**kbartlett** · October 4th 2008, 05:26 AM

If u sub z into $x^4-4x^2+16$ u get $z^2-4z+16$ because $z^2=x^4$ .
$z^2-4z+16$ factorises into $(z-2-i\sqrt12)(z-2+i\sqrt12)$

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

So A=C=0
B= $-2-i\sqrt12$
and D= $-2+i\sqrt12$

**oxrigby** · October 4th 2008, 05:51 AM

Question says,'where ABCD are real constants'''.??

**kbartlett** · October 4th 2008, 06:23 AM

Originally Posted by oxrigby

Question says,'where ABCD are real constants'''.??

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

$(x^2-2-i\sqrt12)(x^2-2+i\sqrt12)$ = $(x^2+4x-\sqrt12)(x^2+4x+\sqrt12)$ = $x^4-4x^2+16$

**Opalg** · October 4th 2008, 06:54 AM

Originally Posted by oxrigby

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

Difference of two squares:

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

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