Solve 2y^2+8y

$$2 y ^ { 2 } + 8 y$$

Factor

$$2y\left(y+4\right)$$

Solution Steps

Factor out 2.

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

Consider y^{2}+4y. Factor out y.

$$y\left(y+4\right)$$

Rewrite the complete factored expression.

$$2y\left(y+4\right)$$

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.

$$2y^{2}+8y=0$$

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.

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

Take the square root of 8^{2}.

$$y=\frac{-8±8}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{-8±8}{4}$$

Now solve the equation y=\frac{-8±8}{4} when ± is plus. Add -8 to 8.

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

Divide 0 by 4.

$$y=0$$

Now solve the equation y=\frac{-8±8}{4} when ± is minus. Subtract 8 from -8.

$$y=-\frac{16}{4}$$

Divide -16 by 4.

$$y=-4$$

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 -4 for x_{2}.

$$2y^{2}+8y=2y\left(y-\left(-4\right)\right)$$

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

$$2y^{2}+8y=2y\left(y+4\right)$$

Evaluate

$$2y\left(y+4\right)$$