Solve 3x+6x^2

$$3x+6 { x }^{ 2 }$$

Factor

$$3x\left(2x+1\right)$$

Solution Steps

Factor out 3.

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

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

$$x\left(1+2x\right)$$

Rewrite the complete factored expression.

$$3x\left(2x+1\right)$$

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.

$$6x^{2}+3x=0$$

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.

$$x=\frac{-3±\sqrt{3^{2}}}{2\times 6}$$

Take the square root of 3^{2}.

$$x=\frac{-3±3}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-3±3}{12}$$

Now solve the equation x=\frac{-3±3}{12} when ± is plus. Add -3 to 3.

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

Divide 0 by 12.

$$x=0$$

Now solve the equation x=\frac{-3±3}{12} when ± is minus. Subtract 3 from -3.

$$x=-\frac{6}{12}$$

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

$$x=-\frac{1}{2}$$

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{1}{2} for x_{2}.

$$6x^{2}+3x=6x\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$6x^{2}+3x=6x\left(x+\frac{1}{2}\right)$$

Add \frac{1}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+3x=6x\times \frac{2x+1}{2}$$

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

$$6x^{2}+3x=3x\left(2x+1\right)$$

Evaluate

$$3x\left(2x+1\right)$$