Solve 2x^2-5x-18

$$2 { x }^{ 2 } -5x-18$$

Factor

$$\left(2x-9\right)\left(x+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-18. To find a and b, set up a system to be solved.

$$a+b=-5 ab=2\left(-18\right)=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$a=-9 b=4$$

Rewrite 2x^{2}-5x-18 as \left(2x^{2}-9x\right)+\left(4x-18\right).

$$\left(2x^{2}-9x\right)+\left(4x-18\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(2x-9\right)+2\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(x+2\right)$$

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.

$$2x^{2}-5x-18=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\left(-18\right)}}{2\times 2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-18\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-18\right)}}{2\times 2}$$

Multiply -8 times -18.

$$x=\frac{-\left(-5\right)±\sqrt{25+144}}{2\times 2}$$

Add 25 to 144.

$$x=\frac{-\left(-5\right)±\sqrt{169}}{2\times 2}$$

Take the square root of 169.

$$x=\frac{-\left(-5\right)±13}{2\times 2}$$

The opposite of -5 is 5.

$$x=\frac{5±13}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{5±13}{4}$$

Now solve the equation x=\frac{5±13}{4} when ± is plus. Add 5 to 13.

$$x=\frac{18}{4}$$

Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{2}$$

Now solve the equation x=\frac{5±13}{4} when ± is minus. Subtract 13 from 5.

$$x=-\frac{8}{4}$$

Divide -8 by 4.

$$x=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9}{2} for x_{1} and -2 for x_{2}.

$$2x^{2}-5x-18=2\left(x-\frac{9}{2}\right)\left(x-\left(-2\right)\right)$$

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

$$2x^{2}-5x-18=2\left(x-\frac{9}{2}\right)\left(x+2\right)$$

Subtract \frac{9}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2x^{2}-5x-18=2\times \frac{2x-9}{2}\left(x+2\right)$$

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

$$2x^{2}-5x-18=\left(2x-9\right)\left(x+2\right)$$

Evaluate

$$\left(2x-9\right)\left(x+2\right)$$