Solve 21c^2+14c

$$21 c ^ { 2 } + 14 c$$

Factor

$$7c\left(3c+2\right)$$

Solution Steps

Factor out 7.

$$7\left(3c^{2}+2c\right)$$

Consider 3c^{2}+2c. Factor out c.

$$c\left(3c+2\right)$$

Rewrite the complete factored expression.

$$7c\left(3c+2\right)$$

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.

$$21c^{2}+14c=0$$

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.

$$c=\frac{-14±\sqrt{14^{2}}}{2\times 21}$$

Take the square root of 14^{2}.

$$c=\frac{-14±14}{2\times 21}$$

Multiply 2 times 21.

$$c=\frac{-14±14}{42}$$

Now solve the equation c=\frac{-14±14}{42} when ± is plus. Add -14 to 14.

$$c=\frac{0}{42}$$

Divide 0 by 42.

$$c=0$$

Now solve the equation c=\frac{-14±14}{42} when ± is minus. Subtract 14 from -14.

$$c=-\frac{28}{42}$$

Reduce the fraction \frac{-28}{42} to lowest terms by extracting and canceling out 14.

$$c=-\frac{2}{3}$$

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{2}{3} for x_{2}.

$$21c^{2}+14c=21c\left(c-\left(-\frac{2}{3}\right)\right)$$

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

$$21c^{2}+14c=21c\left(c+\frac{2}{3}\right)$$

Add \frac{2}{3} to c by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$21c^{2}+14c=21c\times \frac{3c+2}{3}$$

Cancel out 3, the greatest common factor in 21 and 3.

$$21c^{2}+14c=7c\left(3c+2\right)$$

Evaluate

$$7c\left(3c+2\right)$$