Solve (6y^2+10y)

$$( 6 y ^ { 2 } + 10 y )$$

Factor

$$2y\left(3y+5\right)$$

Solution Steps

Factor out 2.

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

Consider 3y^{2}+5y. Factor out y.

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

Rewrite the complete factored expression.

$$2y\left(3y+5\right)$$

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.

$$6y^{2}+10y=0$$

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.

$$y=\frac{-10±\sqrt{10^{2}}}{2\times 6}$$

Take the square root of 10^{2}.

$$y=\frac{-10±10}{2\times 6}$$

Multiply 2 times 6.

$$y=\frac{-10±10}{12}$$

Now solve the equation y=\frac{-10±10}{12} when ± is plus. Add -10 to 10.

$$y=\frac{0}{12}$$

Divide 0 by 12.

$$y=0$$

Now solve the equation y=\frac{-10±10}{12} when ± is minus. Subtract 10 from -10.

$$y=-\frac{20}{12}$$

Reduce the fraction \frac{-20}{12} to lowest terms by extracting and canceling out 4.

$$y=-\frac{5}{3}$$

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{5}{3} for x_{2}.

$$6y^{2}+10y=6y\left(y-\left(-\frac{5}{3}\right)\right)$$

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

$$6y^{2}+10y=6y\left(y+\frac{5}{3}\right)$$

Add \frac{5}{3} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6y^{2}+10y=6y\times \frac{3y+5}{3}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6y^{2}+10y=2y\left(3y+5\right)$$

Evaluate

$$2y\left(3y+5\right)$$