Solve 10c+8c^2

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

Factor

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

Solution Steps

Factor out 2.

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Take the square root of 10^{2}.

$$c=\frac{-10±10}{2\times 8}$$

Multiply 2 times 8.

$$c=\frac{-10±10}{16}$$

Now solve the equation c=\frac{-10±10}{16} when ± is plus. Add -10 to 10.

$$c=\frac{0}{16}$$

Divide 0 by 16.

$$c=0$$

Now solve the equation c=\frac{-10±10}{16} when ± is minus. Subtract 10 from -10.

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

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

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

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{5}{4} for x_{2}.

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

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

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

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

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

Cancel out 4, the greatest common factor in 8 and 4.

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

Evaluate

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