Solve 15y^2+5y

$$15 y ^ { 2 } + 5 y$$

Factor

$$5y\left(3y+1\right)$$

Solution Steps

Factor out 5.

$$5\left(3y^{2}+y\right)$$

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

$$y\left(3y+1\right)$$

Rewrite the complete factored expression.

$$5y\left(3y+1\right)$$

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.

$$15y^{2}+5y=0$$

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.

$$y=\frac{-5±\sqrt{5^{2}}}{2\times 15}$$

Take the square root of 5^{2}.

$$y=\frac{-5±5}{2\times 15}$$

Multiply 2 times 15.

$$y=\frac{-5±5}{30}$$

Now solve the equation y=\frac{-5±5}{30} when ± is plus. Add -5 to 5.

$$y=\frac{0}{30}$$

Divide 0 by 30.

$$y=0$$

Now solve the equation y=\frac{-5±5}{30} when ± is minus. Subtract 5 from -5.

$$y=-\frac{10}{30}$$

Reduce the fraction \frac{-10}{30} to lowest terms by extracting and canceling out 10.

$$y=-\frac{1}{3}$$

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 -\frac{1}{3} for x_{2}.

$$15y^{2}+5y=15y\left(y-\left(-\frac{1}{3}\right)\right)$$

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

$$15y^{2}+5y=15y\left(y+\frac{1}{3}\right)$$

Add \frac{1}{3} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15y^{2}+5y=15y\times \frac{3y+1}{3}$$

Cancel out 3, the greatest common factor in 15 and 3.

$$15y^{2}+5y=5y\left(3y+1\right)$$

Evaluate

$$5y\left(3y+1\right)$$