Solve 6v^2+9v

$$6 v ^ { 2 } + 9 v$$

Factor

$$3v\left(2v+3\right)$$

Solution Steps

Factor out 3.

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

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

$$v\left(2v+3\right)$$

Rewrite the complete factored expression.

$$3v\left(2v+3\right)$$

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.

$$6v^{2}+9v=0$$

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.

$$v=\frac{-9±\sqrt{9^{2}}}{2\times 6}$$

Take the square root of 9^{2}.

$$v=\frac{-9±9}{2\times 6}$$

Multiply 2 times 6.

$$v=\frac{-9±9}{12}$$

Now solve the equation v=\frac{-9±9}{12} when ± is plus. Add -9 to 9.

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

Divide 0 by 12.

$$v=0$$

Now solve the equation v=\frac{-9±9}{12} when ± is minus. Subtract 9 from -9.

$$v=-\frac{18}{12}$$

Reduce the fraction \frac{-18}{12} to lowest terms by extracting and canceling out 6.

$$v=-\frac{3}{2}$$

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

$$6v^{2}+9v=6v\left(v-\left(-\frac{3}{2}\right)\right)$$

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

$$6v^{2}+9v=6v\left(v+\frac{3}{2}\right)$$

Add \frac{3}{2} to v by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6v^{2}+9v=6v\times \frac{2v+3}{2}$$

Cancel out 2, the greatest common factor in 6 and 2.

$$6v^{2}+9v=3v\left(2v+3\right)$$

Evaluate

$$3v\left(2v+3\right)$$