Solve a-12+6a^2

$$a - 12 + 6 a ^ { 2 }$$

Factor

$$\left(3a-4\right)\left(2a+3\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$6a^{2}+a-12$$

Factor the expression by grouping. First, the expression needs to be rewritten as 6a^{2}+pa+qa-12. To find p and q, set up a system to be solved.

$$p+q=1 pq=6\left(-12\right)=-72$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 1.

$$p=-8 q=9$$

Rewrite 6a^{2}+a-12 as \left(6a^{2}-8a\right)+\left(9a-12\right).

$$\left(6a^{2}-8a\right)+\left(9a-12\right)$$

Factor out 2a in the first and 3 in the second group.

$$2a\left(3a-4\right)+3\left(3a-4\right)$$

Factor out common term 3a-4 by using distributive property.

$$\left(3a-4\right)\left(2a+3\right)$$

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.

$$6a^{2}+a-12=0$$

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.

$$a=\frac{-1±\sqrt{1^{2}-4\times 6\left(-12\right)}}{2\times 6}$$

Square 1.

$$a=\frac{-1±\sqrt{1-4\times 6\left(-12\right)}}{2\times 6}$$

Multiply -4 times 6.

$$a=\frac{-1±\sqrt{1-24\left(-12\right)}}{2\times 6}$$

Multiply -24 times -12.

$$a=\frac{-1±\sqrt{1+288}}{2\times 6}$$

Add 1 to 288.

$$a=\frac{-1±\sqrt{289}}{2\times 6}$$

Take the square root of 289.

$$a=\frac{-1±17}{2\times 6}$$

Multiply 2 times 6.

$$a=\frac{-1±17}{12}$$

Now solve the equation a=\frac{-1±17}{12} when ± is plus. Add -1 to 17.

$$a=\frac{16}{12}$$

Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.

$$a=\frac{4}{3}$$

Now solve the equation a=\frac{-1±17}{12} when ± is minus. Subtract 17 from -1.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and -\frac{3}{2} for x_{2}.

$$6a^{2}+a-12=6\left(a-\frac{4}{3}\right)\left(a-\left(-\frac{3}{2}\right)\right)$$

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

$$6a^{2}+a-12=6\left(a-\frac{4}{3}\right)\left(a+\frac{3}{2}\right)$$

Subtract \frac{4}{3} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6a^{2}+a-12=6\times \frac{3a-4}{3}\left(a+\frac{3}{2}\right)$$

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

$$6a^{2}+a-12=6\times \frac{3a-4}{3}\times \frac{2a+3}{2}$$

Multiply \frac{3a-4}{3} times \frac{2a+3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6a^{2}+a-12=6\times \frac{\left(3a-4\right)\left(2a+3\right)}{3\times 2}$$

Multiply 3 times 2.

$$6a^{2}+a-12=6\times \frac{\left(3a-4\right)\left(2a+3\right)}{6}$$

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

$$6a^{2}+a-12=\left(3a-4\right)\left(2a+3\right)$$

Evaluate

$$6a^{2}+a-12$$