Solve 8x^2-2x-3

$$8 { x }^{ 2 } -2x-3$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=8\left(-3\right)=-24$$

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 -2.

$$a=-6 b=4$$

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

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

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

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

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

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

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.

$$8x^{2}-2x-3=0$$

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 8\left(-3\right)}}{2\times 8}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 8\left(-3\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-2\right)±\sqrt{4-32\left(-3\right)}}{2\times 8}$$

Multiply -32 times -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+96}}{2\times 8}$$

Add 4 to 96.

$$x=\frac{-\left(-2\right)±\sqrt{100}}{2\times 8}$$

Take the square root of 100.

$$x=\frac{-\left(-2\right)±10}{2\times 8}$$

The opposite of -2 is 2.

$$x=\frac{2±10}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{2±10}{16}$$

Now solve the equation x=\frac{2±10}{16} when ± is plus. Add 2 to 10.

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

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

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

Now solve the equation x=\frac{2±10}{16} when ± is minus. Subtract 10 from 2.

$$x=-\frac{8}{16}$$

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

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

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

$$8x^{2}-2x-3=8\left(x-\frac{3}{4}\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$8x^{2}-2x-3=8\left(x-\frac{3}{4}\right)\left(x+\frac{1}{2}\right)$$

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

$$8x^{2}-2x-3=8\times \frac{4x-3}{4}\left(x+\frac{1}{2}\right)$$

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

$$8x^{2}-2x-3=8\times \frac{4x-3}{4}\times \frac{2x+1}{2}$$

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

$$8x^{2}-2x-3=8\times \frac{\left(4x-3\right)\left(2x+1\right)}{4\times 2}$$

Multiply 4 times 2.

$$8x^{2}-2x-3=8\times \frac{\left(4x-3\right)\left(2x+1\right)}{8}$$

Cancel out 8, the greatest common factor in 8 and 8.

$$8x^{2}-2x-3=\left(4x-3\right)\left(2x+1\right)$$

Evaluate

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