Solve 2x^2+13x-24

$$2 { x }^{ 2 } +13x-24$$

Factor

$$\left(2x-3\right)\left(x+8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=13 ab=2\left(-24\right)=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 13.

$$a=-3 b=16$$

Rewrite 2x^{2}+13x-24 as \left(2x^{2}-3x\right)+\left(16x-24\right).

$$\left(2x^{2}-3x\right)+\left(16x-24\right)$$

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

$$x\left(2x-3\right)+8\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(x+8\right)$$

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.

$$2x^{2}+13x-24=0$$

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.

$$x=\frac{-13±\sqrt{13^{2}-4\times 2\left(-24\right)}}{2\times 2}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 2\left(-24\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-13±\sqrt{169-8\left(-24\right)}}{2\times 2}$$

Multiply -8 times -24.

$$x=\frac{-13±\sqrt{169+192}}{2\times 2}$$

Add 169 to 192.

$$x=\frac{-13±\sqrt{361}}{2\times 2}$$

Take the square root of 361.

$$x=\frac{-13±19}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-13±19}{4}$$

Now solve the equation x=\frac{-13±19}{4} when ± is plus. Add -13 to 19.

$$x=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{-13±19}{4} when ± is minus. Subtract 19 from -13.

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

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{3}{2} for x_{1} and -8 for x_{2}.

$$2x^{2}+13x-24=2\left(x-\frac{3}{2}\right)\left(x-\left(-8\right)\right)$$

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

$$2x^{2}+13x-24=2\left(x-\frac{3}{2}\right)\left(x+8\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2x^{2}+13x-24=2\times \frac{2x-3}{2}\left(x+8\right)$$

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

$$2x^{2}+13x-24=\left(2x-3\right)\left(x+8\right)$$

Evaluate

$$\left(2x-3\right)\left(x+8\right)$$