Solve 2x^2+15x-8

$$2 { x }^{ 2 } +15x-8$$

Factor

$$\left(2x-1\right)\left(x+8\right)$$

Solution Steps

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

$$a+b=15 ab=2\left(-8\right)=-16$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -16.

$$-1,16 -2,8 -4,4$$

Calculate the sum for each pair.

$$-1+16=15 -2+8=6 -4+4=0$$

The solution is the pair that gives sum 15.

$$a=-1 b=16$$

Rewrite 2x^{2}+15x-8 as \left(2x^{2}-x\right)+\left(16x-8\right).

$$\left(2x^{2}-x\right)+\left(16x-8\right)$$

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

$$x\left(2x-1\right)+8\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(x+8\right)$$

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.

$$2x^{2}+15x-8=0$$

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.

$$x=\frac{-15±\sqrt{15^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\times 2\left(-8\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-15±\sqrt{225-8\left(-8\right)}}{2\times 2}$$

Multiply -8 times -8.

$$x=\frac{-15±\sqrt{225+64}}{2\times 2}$$

Add 225 to 64.

$$x=\frac{-15±\sqrt{289}}{2\times 2}$$

Take the square root of 289.

$$x=\frac{-15±17}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-15±17}{4}$$

Now solve the equation x=\frac{-15±17}{4} when ± is plus. Add -15 to 17.

$$x=\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-15±17}{4} when ± is minus. Subtract 17 from -15.

$$x=-\frac{32}{4}$$

Divide -32 by 4.

$$x=-8$$

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

$$2x^{2}+15x-8=2\left(x-\frac{1}{2}\right)\left(x-\left(-8\right)\right)$$

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

$$2x^{2}+15x-8=2\left(x-\frac{1}{2}\right)\left(x+8\right)$$

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

$$2x^{2}+15x-8=2\times \frac{2x-1}{2}\left(x+8\right)$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2x^{2}+15x-8=\left(2x-1\right)\left(x+8\right)$$

Evaluate

$$\left(2x-1\right)\left(x+8\right)$$