1. Directrix

a^2 - a - 1) ( a^2 +a -1)

2. ash2326

We've got $a^4-3a^2+1$ let's substitute a^2=u so a^4=u^2 now we have $u^2-3u+1$ now we'll use quadratic formula to figure out the factors for $ax^2+bx+c$ the roots are given as $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ here we have u^2-3u+1=0 a=1, b=-3 and c=1 so $u=\frac{3\pm\sqrt{3^2-4*1*1}}{2*1}$ or $u=\frac{3\pm\sqrt{5}}{2}$ so we'll have the factors as $(u-(\frac{3+\sqrt{5}}{2}))(u-(\frac{3-\sqrt{5}}{2}))$ now we have u=a^2 , so the factors are $(a^2-(\frac{3+\sqrt{5}}{2}))(a^2-(\frac{3-\sqrt{5}}{2}))$

3. anonymous

$a^4 - 3a^2 + 1 = (a^2-1)^2 -a^2 =\left(a^2-1+a\right) \left(a^2-1-a\right)$

4. anonymous

Got it. Thanks a lot.

5. Directrix

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