Factor a difficult polynomial $(x^2+16x)^2 = 17^2$ from Apostol's Calculus I

Question

In Apostol's Calculus Vol 1, he asks for the set of all possible values of $x$:

$\{x \,|\, (x^2+16x)^2 = 17^2 \}$

Here is the solution:

$\{1, -17, -8+ \sqrt {47}, -8 - \sqrt {47} \}$

The $1$ and the $-17$ I get, but then I'm left with factoring $x^2+16+ 17 = 0$, which indeed is $-8+ \sqrt {47}$ and $-8 - \sqrt {47}$. However, I'm not sure if this is "hand-calculatable" or would require MatLab or Octave to derive. Is $x^2+16+ 17 = 0$ in fact manually calculatable $-$ and I was sleeping that day in Algebra?

Answer 1

If $(x^2 + 16x)^2 = 17^2$, then $$\begin{align*} 0 &= (x^2 + 16x)^2 - 17^2 \\ &= ((x^2 + 16x) - 17)((x^2 + 16x) + 17) \tag{$1$} \\ &= (x - 1)(x + 17)(x^2 + 16x + 17) \\ &= (x - 1)(x + 17)((x^2 + 16x + 64) - 47) \tag{$2$}\\ &= (x - 1)(x + 17)\left((x+8)^2 - \left(\sqrt{47}\right)^2\right) \\ &= (x - 1)(x + 17)\left(x + 8 - \sqrt{47}\right)\left(x + 8 + \sqrt{47}\right), \tag{$3$} \end{align*}$$ where $(1)$ and $(3)$ follow from the previous step via the difference of squares formula $$a^2 - b^2 = (a-b)(a+b),$$ and the choice to write $17 = 64 - 47$ in step $(2)$ is made in order to complete the square. All that remains is to read off the roots arising from linear factors of the form $(x - r_i)$: $$x \in \left\{ 1, -17, -8 + \sqrt{47}, -8 - \sqrt{47}\right\}.$$

Answer 2

These can be solved using the Quadratic formula.

If $(x^2+16x)^2 = 17^2$, then either $x^2+16x=17$ or $x^2+16x=-17$.

In the first case, you need to solve $x^2+16x-17=0$. From the Quadratic formula, and using the fact that $$16^2 = (15+1)^2 = 225 + 30 + 1 = 256$$ we have $$\begin{align*} x&= \frac{-16 \pm \sqrt{16^2 - 4(-17)}}{2}\\ &= \frac{-16\pm\sqrt{256 + 68}}{2}\\ &= \frac{-16\pm\sqrt{324}}{2}\\ &= -8 \pm \sqrt{\frac{324}{4}}\\ &= -8 \pm \sqrt{81}\\ &= -8\pm 9. \end{align*}$$ This gives you two solutions: $x=-17$ and $x=1$.

The second equation gives

$$\begin{align*} x&= \frac{-16 \pm \sqrt{16^2 - 4(17)}}{2}\\ &= \frac{-16\pm\sqrt{256 - 68}}{2}\\ &= \frac{-16\pm\sqrt{188}}{2}\\ &= -8 \pm \sqrt{\frac{188}{4}}\\ &= -8 \pm \sqrt{47}. \end{align*}$$

This gives the two other solutions, $-8+\sqrt{47}$ and $-8-\sqrt{47}$.

In both calculations, the next-to-last step is taken by breaking up the fraction into two fractions with common denominator $2$; and putting the $2$ into the square root by writing it as $\sqrt{4}$.

In the alternative, one could note that $324 = 4\times 81 = 2^2\times 4^2$ to solve the first one; and that $188=4\times 47$ in the second.

Answer 3

Here is to factorize the second function by hand. Regroup the terms,

$$x^2+16x+ 17 = (x^2+16x+ 64) - 47 =(x+8)^2-(\sqrt{47})^2 $$

Then, apply the identity $a^2-b^2=(a+b)(a-b)$,

$$x^2+16x+ 17 = (x+8+\sqrt{47})(x+8-\sqrt{47})$$