# How do you factor f(x) = x^4 + 4x^3 - 3x^2 +40x + 208?

Jun 14, 2018

${\left(x + 4\right)}^{2} \left({x}^{2} - 4 x + 13\right)$

#### Explanation:

Note that the constant term of the expression is the product of the four roots. So if they are integers, then they must divide into 208. Possible candidates, plus or minus: 1, 2, 4, 8, 13, 16, 26, 52, 104, 208.

Roots of the quartic

A little quick exploration with a calculator reveals that -4 is a root, with $f \left(- 4\right) = 0$. So let's factor that out by algebraic long division:

Divide $x + 4$ into ${x}^{4} + 4 {x}^{3} - 3 {x}^{2} + 40 x + 208$

$x$ goes into ${x}^{4}$ ${x}^{3}$ times, so the remainder is
$\left({x}^{4} + 4 {x}^{3} - 3 {x}^{2} + 40 x + 208\right) - {x}^{3} \left(x + 4\right) =$
${x}^{4} + 4 {x}^{3} - 3 {x}^{2} + 40 x + 208 - {x}^{4} - 4 {x}^{3} =$
$- 3 {x}^{2} + 40 x + 208$

$x$ goes into $- 3 {x}^{2}$ $- 3 x$ times, so the remainder is
$\left(- 3 {x}^{2} + 40 x + 208\right) + 3 x \left(x + 4\right) =$
$- 3 {x}^{2} + 40 x + 208 + 3 {x}^{2} + 12 x =$
$52 x + 208$, which is $x + 4$ multiplied by 52.

So, collecting terms, we have
${x}^{4} + 4 {x}^{3} - 3 {x}^{2} + 40 x + 208 = \left(x + 4\right) \left({x}^{3} - 3 x + 52\right)$
which is a start on a factorisation, with one of the four possible terms found.

Roots of the cubic

Now we have reduced the problem from a quartic (fourth degree) to a cubic (third degree), which is in general a great improvement, though not to the point where it's an easy task to solve in general.

As before, the product of the three remaining roots is the constant coefficient, so if they are integers then they must divide into 52. Possible candidates, plus or minus: 1, 2, 4, 13, 26, 52, a rather shorter list than before.

As before, a little exploration with a calculator reveals that -4 is a root of this equation also. So we factor out another $x + 4$ from the expression.

Divide $x + 4$ into ${x}^{3} - 3 x + 52$

$x$ goes ${x}^{3}$ ${x}^{2}$ times, so the remainder is
(x^3-3x+52)-x^2(x+4))=
${x}^{3} - 3 x + 52 - {x}^{3} - 4 {x}^{2} =$
$- 4 {x}^{2} - 3 x + 52$

$x$ goes into $- 4 {x}^{2}$ $- 4 x$ times, so the remainder is
$\left(- 4 {x}^{2} - 3 x + 52\right) + 4 x \left(x + 4\right) =$
$- 4 {x}^{2} - 3 x + 52 + 4 {x}^{2} + 16 x =$
$13 x + 52$, which is $x + 4$ multiplied by 13.

So, collecting terms, we have
${x}^{4} + 4 {x}^{3} - 3 {x}^{2} + 40 x + 208 = {\left(x + 4\right)}^{2} \left({x}^{2} - 4 x + 13\right)$
with two of the four possible terms found, both equal.

Roots of the quadratic

Now we reach a point where we can solve the equation directly, using the quadratic formula.

${x}^{2} - 4 x + 13 = 0 \Rightarrow x = \frac{1}{2} \left(4 \pm \sqrt{- 36}\right)$

We see here from the square root term that there are no more real roots to be found. So the full solution in terms of integer real roots is as above:

${\left(x + 4\right)}^{2} \left({x}^{2} - 4 x + 13\right)$

We can factorise further in terms of complex roots, and these have integer real and imaginary coefficients, we see from the above:

$x = \frac{1}{2} \left(4 \pm \sqrt{- 36}\right) = \frac{1}{2} \left(4 \pm 6 i\right) = 2 \pm 3 i$

So the full complex factorisation of this expression is

${\left(x + 4\right)}^{2} \left(x - 2 - 3 i\right) \left(x - 2 + 3 i\right)$