Solve 10c^2+25c

$$10 c ^ { 2 } + 25 c$$

Factor

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

Solution Steps

Factor out 5.

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

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

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

Rewrite the complete factored expression.

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

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.

$$10c^{2}+25c=0$$

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.

$$c=\frac{-25±\sqrt{25^{2}}}{2\times 10}$$

Take the square root of 25^{2}.

$$c=\frac{-25±25}{2\times 10}$$

Multiply 2 times 10.

$$c=\frac{-25±25}{20}$$

Now solve the equation c=\frac{-25±25}{20} when ± is plus. Add -25 to 25.

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

Divide 0 by 20.

$$c=0$$

Now solve the equation c=\frac{-25±25}{20} when ± is minus. Subtract 25 from -25.

$$c=-\frac{50}{20}$$

Reduce the fraction \frac{-50}{20} to lowest terms by extracting and canceling out 10.

$$c=-\frac{5}{2}$$

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}{2} for x_{2}.

$$10c^{2}+25c=10c\left(c-\left(-\frac{5}{2}\right)\right)$$

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

$$10c^{2}+25c=10c\left(c+\frac{5}{2}\right)$$

Add \frac{5}{2} to c by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10c^{2}+25c=10c\times \frac{2c+5}{2}$$

Cancel out 2, the greatest common factor in 10 and 2.

$$10c^{2}+25c=5c\left(2c+5\right)$$

Evaluate

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