Solve 15x^2+5x=

$$15 x ^ { 2 } + 5 x =$$

Factor

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

Solution Steps

Factor out 5.

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Take the square root of 5^{2}.

$$x=\frac{-5±5}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-5±5}{30}$$

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

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

Divide 0 by 30.

$$x=0$$

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

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

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

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

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}.

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

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

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

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

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

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

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

Evaluate

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