# How do you long divide 6x^3 + 3x^2 - 4x + 7 by x^2+1?

Jun 15, 2015

$6 x + 3 + \frac{- 10 x + 4}{{x}^{2} + 1}$

#### Explanation:

Let's write
$A \left(x\right) = 6 {x}^{3} + 3 {x}^{2} - 4 x + 7$ and
$B \left(x\right) = {x}^{2} + 1$

We use the rule: we find the first term dividing the term with the highest degree in $A \left(x\right)$ by the term with the highest degree in $B \left(x\right)$, let's call it ${q}_{3 - 2} {x}^{3 - 2}$ then we calculate ${P}_{1} \left(x\right) = A \left(x\right) - B \left(x\right) \cdot {Q}_{3 - 2} \left(x\right)$, which has degree 2

$\frac{6 {x}^{3}}{x} ^ 2 = 6 x \implies {P}_{1} \left(x\right) = 6 {x}^{3} + 3 {x}^{2} - 4 x + 7 - 6 x \left({x}^{2} + 1\right) = 3 {x}^{2} - 10 x + 7$

Now we consider the same process with ${P}_{1} \left(x\right)$, and we have ${q}_{3 - 2 - 1} {x}^{3 - 2 - 1}$ and $R \left(x\right)$.

$\frac{3 {x}^{2}}{x} ^ 2 = 3 \implies R \left(x\right) = 3 {x}^{2} - 10 x + 7 - 3 \left({x}^{2} + 1\right) = - 10 x + 4$

Notice that

$\partial \left(R \left(x\right)\right) < \partial \left(B \left(x\right)\right)$

so we're done, and $Q \left(x\right) = {q}_{3 - 2} {x}^{3 - 2} + {q}_{3 - 2 - 1} {x}^{3} - 2 - 1 = 6 x + 3$ is the quotient and $R \left(x\right)$ is the reminder

NB: I consider $\partial \left(P\right)$ as the degree of $P$