Solve 6x^2-5x-4=

$$6 { x }^{ 2 } -5x-4 =$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-5 ab=6\left(-4\right)=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b 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.

$$a=-8 b=3$$

Rewrite 6x^{2}-5x-4 as \left(6x^{2}-8x\right)+\left(3x-4\right).

$$\left(6x^{2}-8x\right)+\left(3x-4\right)$$

Factor out 2x in 6x^{2}-8x.

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

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

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

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.

$$6x^{2}-5x-4=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 6\left(-4\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-5\right)±\sqrt{25-24\left(-4\right)}}{2\times 6}$$

Multiply -24 times -4.

$$x=\frac{-\left(-5\right)±\sqrt{25+96}}{2\times 6}$$

Add 25 to 96.

$$x=\frac{-\left(-5\right)±\sqrt{121}}{2\times 6}$$

Take the square root of 121.

$$x=\frac{-\left(-5\right)±11}{2\times 6}$$

The opposite of -5 is 5.

$$x=\frac{5±11}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{5±11}{12}$$

Now solve the equation x=\frac{5±11}{12} when ± is plus. Add 5 to 11.

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

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

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

Now solve the equation x=\frac{5±11}{12} when ± is minus. Subtract 11 from 5.

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

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

$$6x^{2}-5x-4=6\left(x-\frac{4}{3}\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$6x^{2}-5x-4=6\left(x-\frac{4}{3}\right)\left(x+\frac{1}{2}\right)$$

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

$$6x^{2}-5x-4=6\times \frac{3x-4}{3}\left(x+\frac{1}{2}\right)$$

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

$$6x^{2}-5x-4=6\times \frac{3x-4}{3}\times \frac{2x+1}{2}$$

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

$$6x^{2}-5x-4=6\times \frac{\left(3x-4\right)\left(2x+1\right)}{3\times 2}$$

Multiply 3 times 2.

$$6x^{2}-5x-4=6\times \frac{\left(3x-4\right)\left(2x+1\right)}{6}$$

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

$$6x^{2}-5x-4=\left(3x-4\right)\left(2x+1\right)$$

Evaluate

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