Solve 10c^2+8c

$$10 c ^ { 2 } + 8 c$$

Factor

$$2c\left(5c+4\right)$$

Solution Steps

Factor out 2.

$$2\left(5c^{2}+4c\right)$$

Consider 5c^{2}+4c. Factor out c.

$$c\left(5c+4\right)$$

Rewrite the complete factored expression.

$$2c\left(5c+4\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$10c^{2}+8c=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$c=\frac{-8±\sqrt{8^{2}}}{2\times 10}$$

Take the square root of 8^{2}.

$$c=\frac{-8±8}{2\times 10}$$

Multiply 2 times 10.

$$c=\frac{-8±8}{20}$$

Now solve the equation c=\frac{-8±8}{20} when ± is plus. Add -8 to 8.

$$c=\frac{0}{20}$$

Divide 0 by 20.

$$c=0$$

Now solve the equation c=\frac{-8±8}{20} when ± is minus. Subtract 8 from -8.

$$c=-\frac{16}{20}$$

Reduce the fraction \frac{-16}{20} to lowest terms by extracting and canceling out 4.

$$c=-\frac{4}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and -\frac{4}{5} for x_{2}.

$$10c^{2}+8c=10c\left(c-\left(-\frac{4}{5}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$10c^{2}+8c=10c\left(c+\frac{4}{5}\right)$$

Add \frac{4}{5} to c by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10c^{2}+8c=10c\times \frac{5c+4}{5}$$

Cancel out 5, the greatest common factor in 10 and 5.

$$10c^{2}+8c=2c\left(5c+4\right)$$

Evaluate

$$2c\left(5c+4\right)$$