Problem source

Given that 
\[ 
x^4 + p x^2 + q x + r = ( x^2 - a x + b ) ( x^2 + a x + c ) , 
\]  
express  $p$, $q$ and $r$ in terms of $a$, $b$ and $c$. 
 
Show also that $ a^2$ is a root of the cubic equation 
$$ 
u^3 + 2 p u^2 + ( p^2 - 4 r ) u - q^2 = 0 .  
$$ 
Explain why this equation  always has a non-negative root,   and verify  that $u = 9$ is a root in the  
case $p = -1$, $q = -6$, $r = 15$ . 
 
Hence, or otherwise, express 
$$y^4 - 8 y^3 + 23 y^2 - 34 y + 39$$ 
as a product of two quadratic factors.

Solution source

\begin{align*}
&& ( x^2 - a x + b ) ( x^2 + a x + c ) &= x^4 + (b+c-a^2)x^2 + a(b-c)x + bc \\
\Rightarrow && x^4 + p x^2 + q x + r &= x^4 + (b+c-a^2)x^2 + a(b-c)x + bc \\
\Rightarrow && p &= b+c-a^2 \tag{1}\\
&& q &= a(b-c) \tag{2}\\
&& r &= bc \tag{3}
\end{align*}

\begin{align*}
(1): && p+a^2 &= b+ c \\
(2): && \frac{q}{a} &= b - c \\
\Rightarrow && b &= \frac12 (p+a^2 + \frac{q}{a}) \\
&& c &= \frac12 (p+a^2 - \frac{q}{a}) \\
(3): && r &=  \frac12 (p+a^2 + \frac{q}{a}) \frac12 (p+a^2 - \frac{q}{a}) \\
\Rightarrow && 4ra^2 &= (pa + a^3 + q)(pa+a^3-q) \\
&&&= (pa+a^3)^2 - q^2 \\
&&&= a^2(p+a^2)^2 -q^2 \\
&&&= a^2(p^2 + 2pa^2 + a^4) - q^2 \\
&&&= pa^2 + 2pa^4 + a^6 - q^2 \\
\end{align*}

Therefore $a^2$ is a root of $u^3 + 2pu^2 + pu - q^2 = 4ru$, ie the given equation.

When $u = 0$, this equation is $-q^2$, therefore the cubic is negative. But as $u \to \infty$ the cubic tends to $\infty$, therefore it must cross the $x$-axis and have a positive root.

If $p=-1, q = -6, r = 15$ then the cubic is:

$u^3 - 2u^2 + (1-60)u -36$ and so when $u = 9$ we have

\begin{align*}
9^3 - 2\cdot 9^2 -59 \cdot 9 -36 &= 9(9^2-2\cdot 9 - 29 -4) \\
&= 9(81 -18-59-4) \\
&= 0
\end{align*}

so $u = 9$ is a root

Let $y=z + 2$
\begin{align*}
&&y^4 - 8 y^3 + 23 y^2 - 34 y + 39 &= (z+2)^4-8(z+2)^3 + 23(z+2)^2 - 34(z+2) + 39 \\
&&&= z^4+8z^3+24z^2+32z+16 - \\
&&&\quad -8z^3-48z^2-96z-64 \\
&&&\quad\quad +23z^2+92z+92 \\
&&&\quad\quad -34z-68 + 39 \\
&&&= z^4-z^2-6z+15
\end{align*}

So conveniently this is $p = -1, q = -6, r = 15$, so we know that $a = 3$ is a sensible thing to true.

$b = \frac12(-1 + 9 + \frac{-6}{3}) = 3$
$c = \frac12(-1+9-\frac{-6}{3}) = 5$

so 

\begin{align*}
&& z^4-z^2-6z+15 &= (z^2-3z+3)(z^2+3z+5) \\
&&y^4 - 8 y^3 + 23 y^2 - 34 y + 39  &= ((y-2)^2-3(y-2)+3)((y-2)^2+3(y-2)+5) \\
&&&= (y^2-4y+4-3y+6+3)(y^2-4y+4+3y-6+5) \\
&&&= (y^2-7y+13)(y^2-y+3)
\end{align*}