Solve 2u^2+6u

$$2 u ^ { 2 } + 6 u$$

Factor

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

Solution Steps

Factor out 2.

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

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

$$u\left(u+3\right)$$

Rewrite the complete factored expression.

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

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.

$$2u^{2}+6u=0$$

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.

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

Take the square root of 6^{2}.

$$u=\frac{-6±6}{2\times 2}$$

Multiply 2 times 2.

$$u=\frac{-6±6}{4}$$

Now solve the equation u=\frac{-6±6}{4} when ± is plus. Add -6 to 6.

$$u=\frac{0}{4}$$

Divide 0 by 4.

$$u=0$$

Now solve the equation u=\frac{-6±6}{4} when ± is minus. Subtract 6 from -6.

$$u=-\frac{12}{4}$$

Divide -12 by 4.

$$u=-3$$

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 -3 for x_{2}.

$$2u^{2}+6u=2u\left(u-\left(-3\right)\right)$$

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

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

Evaluate

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