Solve 8x^2-3x-5

$$8 { x }^{ 2 } -3x-5$$

Factor

$$\left(x-1\right)\left(8x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-3 ab=8\left(-5\right)=-40$$

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 -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -3.

$$a=-8 b=5$$

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

$$\left(8x^{2}-8x\right)+\left(5x-5\right)$$

Factor out 8x in the first and 5 in the second group.

$$8x\left(x-1\right)+5\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(8x+5\right)$$

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.

$$8x^{2}-3x-5=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 8\left(-5\right)}}{2\times 8}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 8\left(-5\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-3\right)±\sqrt{9-32\left(-5\right)}}{2\times 8}$$

Multiply -32 times -5.

$$x=\frac{-\left(-3\right)±\sqrt{9+160}}{2\times 8}$$

Add 9 to 160.

$$x=\frac{-\left(-3\right)±\sqrt{169}}{2\times 8}$$

Take the square root of 169.

$$x=\frac{-\left(-3\right)±13}{2\times 8}$$

The opposite of -3 is 3.

$$x=\frac{3±13}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{3±13}{16}$$

Now solve the equation x=\frac{3±13}{16} when ± is plus. Add 3 to 13.

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

Divide 16 by 16.

$$x=1$$

Now solve the equation x=\frac{3±13}{16} when ± is minus. Subtract 13 from 3.

$$x=-\frac{10}{16}$$

Reduce the fraction \frac{-10}{16} to lowest terms by extracting and canceling out 2.

$$x=-\frac{5}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and -\frac{5}{8} for x_{2}.

$$8x^{2}-3x-5=8\left(x-1\right)\left(x-\left(-\frac{5}{8}\right)\right)$$

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

$$8x^{2}-3x-5=8\left(x-1\right)\left(x+\frac{5}{8}\right)$$

Add \frac{5}{8} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-3x-5=8\left(x-1\right)\times \frac{8x+5}{8}$$

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

$$8x^{2}-3x-5=\left(x-1\right)\left(8x+5\right)$$

Evaluate

$$\left(x-1\right)\left(8x+5\right)$$