Solve 4y^2+20y

$$4 y ^ { 2 } + 20 y$$

Factor

$$4y\left(y+5\right)$$

Solution Steps

Factor out 4.

$$4\left(y^{2}+5y\right)$$

Consider y^{2}+5y. Factor out y.

$$y\left(y+5\right)$$

Rewrite the complete factored expression.

$$4y\left(y+5\right)$$

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.

$$4y^{2}+20y=0$$

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.

$$y=\frac{-20±\sqrt{20^{2}}}{2\times 4}$$

Take the square root of 20^{2}.

$$y=\frac{-20±20}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{-20±20}{8}$$

Now solve the equation y=\frac{-20±20}{8} when ± is plus. Add -20 to 20.

$$y=\frac{0}{8}$$

Divide 0 by 8.

$$y=0$$

Now solve the equation y=\frac{-20±20}{8} when ± is minus. Subtract 20 from -20.

$$y=-\frac{40}{8}$$

Divide -40 by 8.

$$y=-5$$

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 -5 for x_{2}.

$$4y^{2}+20y=4y\left(y-\left(-5\right)\right)$$

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

$$4y^{2}+20y=4y\left(y+5\right)$$

Evaluate

$$4y\left(y+5\right)$$