$$\frac { 8 c - 3 } { 4 c ^ { 2 } - 2 c + 1 } + \frac { 6 } { 8 c ^ { 3 } + 1 } = \frac { 2 } { 2 c + 1 }$$
Solve for c
$c=-\frac{1}{4}=-0.25$
Steps Using Factoring By Grouping
Steps Using the Quadratic Formula
Steps for Completing the Square
Steps Using Factoring By Grouping
Variable $c$ cannot be equal to $-\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $\left(2c+1\right)\left(4c^{2}-2c+1\right)$, the least common multiple of $4c^{2}-2c+1,8c^{3}+1,2c+1$.
Use the distributive property to multiply $4c^{2}-2c+1$ by $2$.
$$16c^{2}+2c+3=8c^{2}-4c+2$$
Subtract $8c^{2}$ from both sides.
$$16c^{2}+2c+3-8c^{2}=-4c+2$$
Combine $16c^{2}$ and $-8c^{2}$ to get $8c^{2}$.
$$8c^{2}+2c+3=-4c+2$$
Add $4c$ to both sides.
$$8c^{2}+2c+3+4c=2$$
Combine $2c$ and $4c$ to get $6c$.
$$8c^{2}+6c+3=2$$
Subtract $2$ from both sides.
$$8c^{2}+6c+3-2=0$$
Subtract $2$ from $3$ to get $1$.
$$8c^{2}+6c+1=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $8c^{2}+ac+bc+1$. To find $a$ and $b$, set up a system to be solved.
$$a+b=6$$ $$ab=8\times 1=8$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $8$.
$$1,8$$ $$2,4$$
Calculate the sum for each pair.
$$1+8=9$$ $$2+4=6$$
The solution is the pair that gives sum $6$.
$$a=2$$ $$b=4$$
Rewrite $8c^{2}+6c+1$ as $\left(8c^{2}+2c\right)+\left(4c+1\right)$.
$$\left(8c^{2}+2c\right)+\left(4c+1\right)$$
Factor out $2c$ in $8c^{2}+2c$.
$$2c\left(4c+1\right)+4c+1$$
Factor out common term $4c+1$ by using distributive property.
$$\left(4c+1\right)\left(2c+1\right)$$
To find equation solutions, solve $4c+1=0$ and $2c+1=0$.
$$c=-\frac{1}{4}$$ $$c=-\frac{1}{2}$$
Variable $c$ cannot be equal to $-\frac{1}{2}$.
$$c=-\frac{1}{4}$$
Steps Using the Quadratic Formula
Variable $c$ cannot be equal to $-\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $\left(2c+1\right)\left(4c^{2}-2c+1\right)$, the least common multiple of $4c^{2}-2c+1,8c^{3}+1,2c+1$.
Use the distributive property to multiply $4c^{2}-2c+1$ by $2$.
$$16c^{2}+2c+3=8c^{2}-4c+2$$
Subtract $8c^{2}$ from both sides.
$$16c^{2}+2c+3-8c^{2}=-4c+2$$
Combine $16c^{2}$ and $-8c^{2}$ to get $8c^{2}$.
$$8c^{2}+2c+3=-4c+2$$
Add $4c$ to both sides.
$$8c^{2}+2c+3+4c=2$$
Combine $2c$ and $4c$ to get $6c$.
$$8c^{2}+6c+3=2$$
Subtract $2$ from both sides.
$$8c^{2}+6c+3-2=0$$
Subtract $2$ from $3$ to get $1$.
$$8c^{2}+6c+1=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $8$ for $a$, $6$ for $b$, and $1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$c=\frac{-6±\sqrt{6^{2}-4\times 8}}{2\times 8}$$
Square $6$.
$$c=\frac{-6±\sqrt{36-4\times 8}}{2\times 8}$$
Multiply $-4$ times $8$.
$$c=\frac{-6±\sqrt{36-32}}{2\times 8}$$
Add $36$ to $-32$.
$$c=\frac{-6±\sqrt{4}}{2\times 8}$$
Take the square root of $4$.
$$c=\frac{-6±2}{2\times 8}$$
Multiply $2$ times $8$.
$$c=\frac{-6±2}{16}$$
Now solve the equation $c=\frac{-6±2}{16}$ when $±$ is plus. Add $-6$ to $2$.
$$c=-\frac{4}{16}$$
Reduce the fraction $\frac{-4}{16}$ to lowest terms by extracting and canceling out $4$.
$$c=-\frac{1}{4}$$
Now solve the equation $c=\frac{-6±2}{16}$ when $±$ is minus. Subtract $2$ from $-6$.
$$c=-\frac{8}{16}$$
Reduce the fraction $\frac{-8}{16}$ to lowest terms by extracting and canceling out $8$.
$$c=-\frac{1}{2}$$
The equation is now solved.
$$c=-\frac{1}{4}$$ $$c=-\frac{1}{2}$$
Variable $c$ cannot be equal to $-\frac{1}{2}$.
$$c=-\frac{1}{4}$$
Steps for Completing the Square
Variable $c$ cannot be equal to $-\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $\left(2c+1\right)\left(4c^{2}-2c+1\right)$, the least common multiple of $4c^{2}-2c+1,8c^{3}+1,2c+1$.
Use the distributive property to multiply $4c^{2}-2c+1$ by $2$.
$$16c^{2}+2c+3=8c^{2}-4c+2$$
Subtract $8c^{2}$ from both sides.
$$16c^{2}+2c+3-8c^{2}=-4c+2$$
Combine $16c^{2}$ and $-8c^{2}$ to get $8c^{2}$.
$$8c^{2}+2c+3=-4c+2$$
Add $4c$ to both sides.
$$8c^{2}+2c+3+4c=2$$
Combine $2c$ and $4c$ to get $6c$.
$$8c^{2}+6c+3=2$$
Subtract $3$ from both sides.
$$8c^{2}+6c=2-3$$
Subtract $3$ from $2$ to get $-1$.
$$8c^{2}+6c=-1$$
Divide both sides by $8$.
$$\frac{8c^{2}+6c}{8}=-\frac{1}{8}$$
Dividing by $8$ undoes the multiplication by $8$.
$$c^{2}+\frac{6}{8}c=-\frac{1}{8}$$
Reduce the fraction $\frac{6}{8}$ to lowest terms by extracting and canceling out $2$.
$$c^{2}+\frac{3}{4}c=-\frac{1}{8}$$
Divide $\frac{3}{4}$, the coefficient of the $x$ term, by $2$ to get $\frac{3}{8}$. Then add the square of $\frac{3}{8}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
Add $-\frac{1}{8}$ to $\frac{9}{64}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
$$c^{2}+\frac{3}{4}c+\frac{9}{64}=\frac{1}{64}$$
Factor $c^{2}+\frac{3}{4}c+\frac{9}{64}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(c+\frac{3}{8}\right)^{2}=\frac{1}{64}$$
Take the square root of both sides of the equation.
