Question

Please help me factor this biquadratic expression: a^4 - 3a^2 + 1

Answer 1

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

Answer 2

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
\[(a^2-(\frac{3+\sqrt{5}}{2}))(a^2-(\frac{3-\sqrt{5}}{2}))\]

Answer 3

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

Answer 4

Got it. Thanks a lot.

Answer 5

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