Solve 4c+6c^2

$$4 c + 6 c ^ { 2 }$$

Factor

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

Solution Steps

Factor out 2.

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

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

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

Rewrite the complete factored expression.

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

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.

$$6c^{2}+4c=0$$

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.

$$c=\frac{-4±\sqrt{4^{2}}}{2\times 6}$$

Take the square root of 4^{2}.

$$c=\frac{-4±4}{2\times 6}$$

Multiply 2 times 6.

$$c=\frac{-4±4}{12}$$

Now solve the equation c=\frac{-4±4}{12} when ± is plus. Add -4 to 4.

$$c=\frac{0}{12}$$

Divide 0 by 12.

$$c=0$$

Now solve the equation c=\frac{-4±4}{12} when ± is minus. Subtract 4 from -4.

$$c=-\frac{8}{12}$$

Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.

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

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}.

$$6c^{2}+4c=6c\left(c-\left(-\frac{2}{3}\right)\right)$$

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

$$6c^{2}+4c=6c\left(c+\frac{2}{3}\right)$$

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

$$6c^{2}+4c=6c\times \frac{3c+2}{3}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6c^{2}+4c=2c\left(3c+2\right)$$

Evaluate

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