# How do I find the partial-fraction decomposition of (2x^3+7x^2-2x+6)/(x^4+4)?

Aug 8, 2015

First factor the denominator to find ${x}^{4} + 4 = \left({x}^{2} + 2 x + 2\right) \left({x}^{2} - 2 x + 2\right)$

Then solve $2 {x}^{3} + 7 {x}^{2} - 2 x + 6 = \left({x}^{2} - 2 x + 2\right) \left(A x + B\right) + \left({x}^{2} + 2 x + 2\right) \left(C x + D\right)$

finding $A = 0$, $B = 3$, $C = 2$ and $D = 0$.

#### Explanation:

${x}^{4} + 4$ has no linear factors with Real coefficients since ${x}^{4} + 4 \ge 4 > 0$ for all $x \in \mathbb{R}$.

It will have quadratic factors with Real coefficients.

${x}^{4} + 4 = \left(a {x}^{2} + b x + c\right) \left({\mathrm{dx}}^{2} + e x + f\right)$

Without loss of generalisation, $a = d = 1$, so

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

Then looking at the coefficient of ${x}^{3}$, we have $e = - b$, so

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

Then looking at the coefficient of $x$ we have $b \left(f - c\right) = 0$, so either $b = 0$ or $c = f$.

If $b = 0$ then $c + f = 0$ and $c f = - {c}^{2} = 4$, which has no Real solutions (it has solutions $c = 2 i$, $f = - 2 i$).

If $c = f$ and $c f = 4$ then $c = f = \pm 2$. Then the coefficient of ${x}^{2}$ tells us $f + c - {b}^{2} = 0$. Now if $b$ is Real, then ${b}^{2} \ge 0$, so $f + c \ge 0$, hence $f = c = 2$. Then ${b}^{2} = 4$, so $b = \pm 2$.

So we have ${x}^{4} + 4 = \left({x}^{2} + 2 x + 2\right) \left({x}^{2} - 2 x + 2\right)$

Now attempt to solve:

$\frac{2 {x}^{3} + 7 {x}^{2} - 2 x + 6}{{x}^{4} + 4} = \frac{A x + B}{{x}^{2} + 2 x + 2} + \frac{C x + D}{{x}^{2} - 2 x + 2}$

If we multiply through by $\left({x}^{4} + 4\right) = \left({x}^{2} + 2 x + 2\right) \left({x}^{2} - 2 x + 2\right)$ then this becomes:

$2 {x}^{3} + 7 {x}^{2} - 2 x + 6$

$= \left({x}^{2} - 2 x + 2\right) \left(A x + B\right) + \left({x}^{2} + 2 x + 2\right) \left(C x + D\right)$

$= \left(A + C\right) {x}^{3} + \left(B - 2 A + D + 2 C\right) {x}^{2} + 2 \left(A - B + C + D\right) x + 2 \left(B + D\right)$

Equating coefficients we get:

(i) $A + C = 2$
(ii) $B - 2 A + D + 2 C = 7$
(iii) $2 \left(A - B + C + D\right) = - 2$
(iv) $2 \left(B + D\right) = 6$

From (i) and (iv) we get:

(v) $C = 2 - A$ and
(vi) $D = 3 - B$

Substitute these into (ii) to get:

$\textcolor{red}{\cancel{\textcolor{b l a c k}{B}}} - 2 A + 3 - \textcolor{red}{\cancel{\textcolor{b l a c k}{B}}} + 4 - 2 A = 7$

Hence $A = 0$. Then from (v) $C = 2$.

Substitute $A = 0$, $C = 2$ and $D = 3 - B$ into (iii) to get:

$2 \left(0 - B + 2 + 3 - B\right) = - 2$

Divide both sides by $2$ to find:

$5 - 2 B = - 1$

Hence $B = 3$. Then from (vi) $D = 0$.

So

$\frac{2 {x}^{3} + 7 {x}^{2} - 2 x + 6}{{x}^{4} + 4} = \frac{3}{{x}^{2} + 2 x + 2} + \frac{2 x}{{x}^{2} - 2 x + 2}$

Aug 9, 2015

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

$\frac{2 {x}^{3} + 7 {x}^{2} - 2 x + 6}{{x}^{4} + 4} = \frac{3}{{x}^{2} + 2 x + 2} + \frac{2 x}{{x}^{2} - 2 x + 2}$

#### Explanation:

Here is a quick way to find the quadratic factors of ${x}^{4} + 4$ using complex arithmetic ...

$\sqrt{- 4} = 2 i$

$\sqrt{2 i} = i + 1$

If ${x}_{1}$ is a root of ${x}^{4} + 4 = 0$, then so are $- {x}_{1}$, $\overline{{x}_{1}}$ and $\overline{- {x}_{1}}$, where $\overline{{x}_{1}}$ means the complex conjugate. This four-fold symmetry occurs due to the only power of $x$ being ${x}^{4}$.

Here are the roots in the complex plane:

graph{((x-1)^2+(y-1)^2 - 0.01)((x+1)^2+(y-1)^2 - 0.01)((x-1)^2+(y+1)^2 - 0.01)((x+1)^2+(y+1)^2-0.01) = 0 [-5, 5, -2.5, 2.5]}

So ${x}^{4} + 4 = \left(x - 1 - i\right) \left(x - 1 + i\right) \left(x + 1 - i\right) \left(x + 1 + i\right)$

Now pick pairs of factors to multiply to get Real coefficients.

To do this, pick the ones which are complex conjugates:

$\left(x + 1 - i\right) \left(x + 1 + i\right) = \left(\left(x + 1\right) - i\right) \left(\left(x + 1\right) + i\right)$

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

and

$\left(x - 1 - i\right) \left(x - 1 + i\right) = \left(\left(x - 1\right) - i\right) \left(\left(x - 1\right) + i\right)$

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

For the rest of the partial fraction decomposition, see the other answers.