Solve 7x^6+x^3+2x+5/x^2=q(x)+r(x)/b(x)

$$\frac { 7 x ^ { 6 } + x ^ { 3 } + 2 x + 5 } { x ^ { 2 } } = q ( x ) + \frac { r ( x ) } { b ( x ) }$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{rx^{2}}{-7x^{6}+qx^{3}-x^{3}-2x-5}\text{, }&x\neq 0\text{ and }r\neq 0\text{ and }x\left(-7x^{5}+qx^{2}-x^{2}-2\right)\neq 5\text{ and }q\neq \frac{7x^{6}+x^{3}+2x+5}{x^{3}}\\b\neq 0\text{, }&x\neq 0\text{ and }q=\frac{7x^{6}+x^{3}+2x+5}{x^{3}}\text{ and }r=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by bx^{2}, the least common multiple of x^{2},bx.

$$b\left(7x^{6}+x^{3}+2x+5\right)=qxbx^{2}+xrx$$

Use the distributive property to multiply b by 7x^{6}+x^{3}+2x+5.

$$7bx^{6}+bx^{3}+2bx+5b=qxbx^{2}+xrx$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$7bx^{6}+bx^{3}+2bx+5b=qx^{3}b+xrx$$

Multiply x and x to get x^{2}.

$$7bx^{6}+bx^{3}+2bx+5b=qx^{3}b+x^{2}r$$

Subtract qx^{3}b from both sides.

$$7bx^{6}+bx^{3}+2bx+5b-qx^{3}b=x^{2}r$$

Reorder the terms.

$$7bx^{6}-bqx^{3}+bx^{3}+2bx+5b=rx^{2}$$

Combine all terms containing b.

$$\left(7x^{6}-qx^{3}+x^{3}+2x+5\right)b=rx^{2}$$

Divide both sides by -qx^{3}+7x^{6}+x^{3}+2x+5.

$$\frac{\left(7x^{6}-qx^{3}+x^{3}+2x+5\right)b}{7x^{6}-qx^{3}+x^{3}+2x+5}=\frac{rx^{2}}{7x^{6}-qx^{3}+x^{3}+2x+5}$$

Dividing by -qx^{3}+7x^{6}+x^{3}+2x+5 undoes the multiplication by -qx^{3}+7x^{6}+x^{3}+2x+5.

$$b=\frac{rx^{2}}{7x^{6}-qx^{3}+x^{3}+2x+5}$$

Variable b cannot be equal to 0.

$$b=\frac{rx^{2}}{7x^{6}-qx^{3}+x^{3}+2x+5}\text{, }b\neq 0$$

Solve for q

$$q=7x^{3}+1-\frac{r}{bx}+\frac{2}{x^{2}}+\frac{5}{x^{3}}$$

$$b\neq 0\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by bx^{2}, the least common multiple of x^{2},bx.

$$b\left(7x^{6}+x^{3}+2x+5\right)=qxbx^{2}+xrx$$

Use the distributive property to multiply b by 7x^{6}+x^{3}+2x+5.

$$7bx^{6}+bx^{3}+2bx+5b=qxbx^{2}+xrx$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$7bx^{6}+bx^{3}+2bx+5b=qx^{3}b+xrx$$

Multiply x and x to get x^{2}.

$$7bx^{6}+bx^{3}+2bx+5b=qx^{3}b+x^{2}r$$

Swap sides so that all variable terms are on the left hand side.

$$qx^{3}b+x^{2}r=7bx^{6}+bx^{3}+2bx+5b$$

Subtract x^{2}r from both sides.

$$qx^{3}b=7bx^{6}+bx^{3}+2bx+5b-x^{2}r$$

Reorder the terms.

$$bqx^{3}=7bx^{6}+bx^{3}-rx^{2}+2bx+5b$$

The equation is in standard form.

$$bx^{3}q=7bx^{6}+bx^{3}-rx^{2}+2bx+5b$$

Divide both sides by bx^{3}.

$$\frac{bx^{3}q}{bx^{3}}=\frac{7bx^{6}+bx^{3}-rx^{2}+2bx+5b}{bx^{3}}$$

Dividing by bx^{3} undoes the multiplication by bx^{3}.

$$q=\frac{7bx^{6}+bx^{3}-rx^{2}+2bx+5b}{bx^{3}}$$

Divide 7bx^{6}+bx^{3}+2bx+5b-x^{2}r by bx^{3}.

$$q=7x^{3}+1-\frac{r}{bx}+\frac{2}{x^{2}}+\frac{5}{x^{3}}$$