Solve 12a^2+3a

$$12 a ^ { 2 } + 3 a$$

Factor

$$3a\left(4a+1\right)$$

Solution Steps

Factor out 3.

$$3\left(4a^{2}+a\right)$$

Consider 4a^{2}+a. Factor out a.

$$a\left(4a+1\right)$$

Rewrite the complete factored expression.

$$3a\left(4a+1\right)$$

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.

$$12a^{2}+3a=0$$

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.

$$a=\frac{-3±\sqrt{3^{2}}}{2\times 12}$$

Take the square root of 3^{2}.

$$a=\frac{-3±3}{2\times 12}$$

Multiply 2 times 12.

$$a=\frac{-3±3}{24}$$

Now solve the equation a=\frac{-3±3}{24} when ± is plus. Add -3 to 3.

$$a=\frac{0}{24}$$

Divide 0 by 24.

$$a=0$$

Now solve the equation a=\frac{-3±3}{24} when ± is minus. Subtract 3 from -3.

$$a=-\frac{6}{24}$$

Reduce the fraction \frac{-6}{24} to lowest terms by extracting and canceling out 6.

$$a=-\frac{1}{4}$$

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}{4} for x_{2}.

$$12a^{2}+3a=12a\left(a-\left(-\frac{1}{4}\right)\right)$$

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

$$12a^{2}+3a=12a\left(a+\frac{1}{4}\right)$$

Add \frac{1}{4} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12a^{2}+3a=12a\times \frac{4a+1}{4}$$

Cancel out 4, the greatest common factor in 12 and 4.

$$12a^{2}+3a=3a\left(4a+1\right)$$

Evaluate

$$3a\left(4a+1\right)$$