# How do you solve (x^2 - 16)(x - 3)^2 + 9x^2 = 0?

Sep 5, 2015

Find:

$\left({x}^{2} - 16\right) {\left(x - 3\right)}^{2} + 9 {x}^{2} = \left({x}^{2} + 2 x - 6\right) \left({x}^{2} - 8 x + 24\right)$

hence Real roots $x = - 1 \pm \sqrt{7}$

#### Explanation:

Let $f \left(x\right) = \left({x}^{2} - 16\right) {\left(x - 3\right)}^{2} + 9 {x}^{2}$

$= \left({x}^{2} - 16\right) \left({x}^{2} - 6 x + 9\right) + 9 {x}^{2}$

$= {x}^{4} - 6 {x}^{3} + 2 {x}^{2} + 96 x - 144$

If $\left\mid x \right\mid \ge 4$ then $\left({x}^{2} - 16\right) \ge 0$, ${\left(x - 3\right)}^{2} > 0$ and $9 {x}^{2} > 0$, so $f \left(x\right) > 0$.

So the only Real roots of $f \left(x\right) = 0$ are in the range $- 4 < x < 4$.

The rational root theorem tells us that any rational roots of $f \left(x\right) = 0$ must be of the form $\frac{p}{q}$ in lowest terms, where $p , q \in \mathbb{Z}$, $q > 0$, $p$ is a divisor of the constant term $144$ and $q$ is a divisor of the coefficient $1$ of the leading term.

So the only possible rational roots are integer divisors of $144$ strictly between $- 4$ and $4$. That gives the only possibilities as $\pm 1$, $\pm 2$, $\pm 3$. None of these work, so $f \left(x\right) = 0$ has no rational roots.

Next, suppose $f \left(x\right) = \left({x}^{2} + a x + b\right) \left({x}^{2} + c x + d\right)$ for some integers $a$, $b$, $c$ and $d$.

What can we find out about these integers?

$\left({x}^{2} + a x + b\right) \left({x}^{2} + c x + d\right) = {x}^{4} + \left(a + c\right) {x}^{3} + \left(b + d + a c\right) {x}^{2} + \left(a d + b c\right) x + b d$

Equating coefficients, we get:

(i) $a + c = - 6$
(ii) $b + d + a c = 2$
(iii) $a d + b c = 96$
(iv) $b d = - 144$

Suppose $a$ is odd.
Then from (i) $c$ is also odd, so $a c$ is odd.
Then from (ii) $b + d$ is odd, so one of $b$ and $d$ is odd and the other even.
Then one of $a d$ and $b c$ is odd and the other even.
So $a d + b c$ is odd.
But (iii) tells us that $a d + b c = 96$ which is even, so we have a contradiction.
Therefore $a$ is even.

Since $a$ is even, then from (i) we get $c$ is even too. Thus $a c$ is a multiple of $4$.
Then from (ii), we find $b + d$ is even but not a multiple of $4$.
From (iv), at least one of $b$ and $d$ is even, hence they both are.

That leaves us with the following possibilities for $\left(b , d\right)$ with $b < d$:

$\left(- 72 , 2\right)$, $\left(- 24 , 6\right)$, $\left(- 18 , 8\right)$, $\left(- 8 , 18\right)$, $\left(- 6 , 24\right)$, $\left(- 2 , 72\right)$

..with sums:

$- 70$, $- 18$, $- 10$, $10$, $18$, $70$

..which from (ii) give us possible values for $a c$:

$72$, $20$, $12$, $- 8$, $- 16$, $- 68$

..which in combination with (i) give us possible integer values of $\left(a , c\right)$:

(none), (none), (none), (none), $\left(- 8 , 2\right)$ / $\left(2 , - 8\right)$, (none)

So $\left(b , d\right) = \left(- 6 , 24\right)$ and $\left(a , c\right) = \left(- 8 , 2\right) \text{ or } \left(2 , - 8\right)$

From (iii) we quickly find:

$\left(a , c\right) = \left(2 , - 8\right)$

Hence:

$f \left(x\right) = \left({x}^{2} + 2 x - 6\right) \left({x}^{2} - 8 x + 24\right)$

Using the quadratic formula, the first of these factors has roots:

$x = - 1 \pm \sqrt{7}$

and the second only has complex roots.