# How do you factor  x^4 + 4x^3 - 22x^2 + 4x + 16?

Aug 28, 2016

${x}^{4} + 4 {x}^{3} - 22 {x}^{2} + 4 x + 16$

$= \left(x + 3 - \sqrt{17}\right) \left(x + 3 + \sqrt{17}\right) \left(x - 1 - \sqrt{3}\right) \left(x - 1 + \sqrt{3}\right)$

#### Explanation:

$f \left(x\right) = {x}^{4} + 4 {x}^{3} - 22 {x}^{2} + 4 x + 16$

We can try the rational roots theorem, which tells us that any rational zeros of $f \left(x\right)$ are expressibl in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $16$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 8 , \pm 16$

None of these works, so $f \left(x\right)$ has no rational zeros and therefore no linear factors with integer coefficients.

Is there a pair of quadratic factors with integer coefficients?

We can simplify the problem by making a linear substitution.

${x}^{4} + 4 {x}^{3} - 22 {x}^{2} + 4 x + 16$

$= {\left(x + 1\right)}^{4} - 28 {\left(x + 1\right)}^{2} + 56 \left(x + 1\right) - 13$

$= {t}^{4} - 28 {t}^{2} + 56 t - 13$

where $t = x + 1$

If this simplified quartic has quadratic factors with integer coefficients then they must take one of the following forms:

• $\left({t}^{2} + a t + 13\right) \left({t}^{2} - a t - 1\right)$

• $\left({t}^{2} + a t - 13\right) \left({t}^{2} - a t + 1\right)$

With a little comparing of coefficients we find:

${t}^{4} - 28 {t}^{2} + 56 t - 13 = \left({t}^{2} + 4 t - 13\right) \left({t}^{2} - 4 t + 1\right)$

Hence:

${x}^{4} + 4 {x}^{3} - 22 {x}^{2} + 4 x + 16$

$= \left({\left(x + 1\right)}^{2} + 4 \left(x + 1\right) - 13\right) \left({\left(x + 1\right)}^{2} - 4 \left(x + 1\right) + 1\right)$

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

$= \left({\left(x + 3\right)}^{2} - 17\right) \left({\left(x - 1\right)}^{2} - 3\right)$

$= \left(x + 3 - \sqrt{17}\right) \left(x + 3 + \sqrt{17}\right) \left(x - 1 - \sqrt{3}\right) \left(x - 1 + \sqrt{3}\right)$