Solve 12a^2-5a-2

$$12 { a }^{ 2 } -5a-2$$

Factor

$$\left(3a-2\right)\left(4a+1\right)$$

Solution Steps

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

$$p+q=-5 pq=12\left(-2\right)=-24$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -5.

$$p=-8 q=3$$

Rewrite 12a^{2}-5a-2 as \left(12a^{2}-8a\right)+\left(3a-2\right).

$$\left(12a^{2}-8a\right)+\left(3a-2\right)$$

Factor out 4a in 12a^{2}-8a.

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

Factor out common term 3a-2 by using distributive property.

$$\left(3a-2\right)\left(4a+1\right)$$

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.

$$12a^{2}-5a-2=0$$

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.

$$a=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 12\left(-2\right)}}{2\times 12}$$

Square -5.

$$a=\frac{-\left(-5\right)±\sqrt{25-4\times 12\left(-2\right)}}{2\times 12}$$

Multiply -4 times 12.

$$a=\frac{-\left(-5\right)±\sqrt{25-48\left(-2\right)}}{2\times 12}$$

Multiply -48 times -2.

$$a=\frac{-\left(-5\right)±\sqrt{25+96}}{2\times 12}$$

Add 25 to 96.

$$a=\frac{-\left(-5\right)±\sqrt{121}}{2\times 12}$$

Take the square root of 121.

$$a=\frac{-\left(-5\right)±11}{2\times 12}$$

The opposite of -5 is 5.

$$a=\frac{5±11}{2\times 12}$$

Multiply 2 times 12.

$$a=\frac{5±11}{24}$$

Now solve the equation a=\frac{5±11}{24} when ± is plus. Add 5 to 11.

$$a=\frac{16}{24}$$

Reduce the fraction \frac{16}{24} to lowest terms by extracting and canceling out 8.

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

Now solve the equation a=\frac{5±11}{24} when ± is minus. Subtract 11 from 5.

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

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 \frac{2}{3} for x_{1} and -\frac{1}{4} for x_{2}.

$$12a^{2}-5a-2=12\left(a-\frac{2}{3}\right)\left(a-\left(-\frac{1}{4}\right)\right)$$

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

$$12a^{2}-5a-2=12\left(a-\frac{2}{3}\right)\left(a+\frac{1}{4}\right)$$

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

$$12a^{2}-5a-2=12\times \frac{3a-2}{3}\left(a+\frac{1}{4}\right)$$

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

$$12a^{2}-5a-2=12\times \frac{3a-2}{3}\times \frac{4a+1}{4}$$

Multiply \frac{3a-2}{3} times \frac{4a+1}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12a^{2}-5a-2=12\times \frac{\left(3a-2\right)\left(4a+1\right)}{3\times 4}$$

Multiply 3 times 4.

$$12a^{2}-5a-2=12\times \frac{\left(3a-2\right)\left(4a+1\right)}{12}$$

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

$$12a^{2}-5a-2=\left(3a-2\right)\left(4a+1\right)$$

Evaluate

$$\left(3a-2\right)\left(4a+1\right)$$