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

Aug 28, 2015

Determine that ${x}^{4} + 3 {x}^{2} + 4$ only has quadratic factors with Real coefficients and solve some simultaneous equations to find:

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

#### Explanation:

${x}^{4} + 3 {x}^{2} + 4$ has no linear factors with Real coefficients since ${x}^{4} + 3 {x}^{2} + 4 \ge 4$ for all $x \in \mathbb{R}$.

So let's look for quadratic factors. Since there's no term in ${x}^{3}$, these must take the form:

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

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

Equating coefficients we find:

(i) $b + c - {a}^{2} = 3$

(ii) $a \left(c - b\right) = 0$

(iii) $b c = 4$

From (ii), $a = 0$ and/or $b = c$.

If $a = 0$, then we are effectively trying to find linear factors of ${t}^{2} + 3 t + 4$, where $t = {x}^{2}$. This quadratic in $t$ has a negative discriminant (${3}^{2} - 4 \cdot 1 \cdot 4 = 9 - 16 = - 7$), so only Complex zeros.

So try $b = c$:

From (iii), we get $b = c = 2$ or $b = c = - 2$. If $a \in \mathbb{R}$ then ${a}^{2} \ge 0$ and from (i) we get $b + c = 3 + {a}^{2} \ge 3$. So $b = c = 2$.

Then ${a}^{2} = b + c - 3 = 2 + 2 - 3 = 1$. So $a = \pm 1$

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