Solve 5y+20y^2

$$5 y + 20 y ^ { 2 }$$

Factor

$$5y\left(4y+1\right)$$

Solution Steps

Factor out 5.

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

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

$$y\left(1+4y\right)$$

Rewrite the complete factored expression.

$$5y\left(4y+1\right)$$

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.

$$20y^{2}+5y=0$$

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.

$$y=\frac{-5±\sqrt{5^{2}}}{2\times 20}$$

Take the square root of 5^{2}.

$$y=\frac{-5±5}{2\times 20}$$

Multiply 2 times 20.

$$y=\frac{-5±5}{40}$$

Now solve the equation y=\frac{-5±5}{40} when ± is plus. Add -5 to 5.

$$y=\frac{0}{40}$$

Divide 0 by 40.

$$y=0$$

Now solve the equation y=\frac{-5±5}{40} when ± is minus. Subtract 5 from -5.

$$y=-\frac{10}{40}$$

Reduce the fraction \frac{-10}{40} to lowest terms by extracting and canceling out 10.

$$y=-\frac{1}{4}$$

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 -\frac{1}{4} for x_{2}.

$$20y^{2}+5y=20y\left(y-\left(-\frac{1}{4}\right)\right)$$

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

$$20y^{2}+5y=20y\left(y+\frac{1}{4}\right)$$

Add \frac{1}{4} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20y^{2}+5y=20y\times \frac{4y+1}{4}$$

Cancel out 4, the greatest common factor in 20 and 4.

$$20y^{2}+5y=5y\left(4y+1\right)$$

Evaluate

$$5y\left(4y+1\right)$$