Solve 12y^2+96y

$$12 y ^ { 2 } + 96 y$$

Factor

$$12y\left(y+8\right)$$

Solution Steps

Factor out 12.

$$12\left(y^{2}+8y\right)$$

Consider y^{2}+8y. Factor out y.

$$y\left(y+8\right)$$

Rewrite the complete factored expression.

$$12y\left(y+8\right)$$

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.

$$12y^{2}+96y=0$$

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.

$$y=\frac{-96±\sqrt{96^{2}}}{2\times 12}$$

Take the square root of 96^{2}.

$$y=\frac{-96±96}{2\times 12}$$

Multiply 2 times 12.

$$y=\frac{-96±96}{24}$$

Now solve the equation y=\frac{-96±96}{24} when ± is plus. Add -96 to 96.

$$y=\frac{0}{24}$$

Divide 0 by 24.

$$y=0$$

Now solve the equation y=\frac{-96±96}{24} when ± is minus. Subtract 96 from -96.

$$y=-\frac{192}{24}$$

Divide -192 by 24.

$$y=-8$$

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 -8 for x_{2}.

$$12y^{2}+96y=12y\left(y-\left(-8\right)\right)$$

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

$$12y^{2}+96y=12y\left(y+8\right)$$

Evaluate

$$12y\left(y+8\right)$$