Solve 2x^2+9x-5

$$2 { x }^{ 2 } +9x-5$$

Factor

$$\left(2x-1\right)\left(x+5\right)$$

Solution Steps

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

$$a+b=9 ab=2\left(-5\right)=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 9.

$$a=-1 b=10$$

Rewrite 2x^{2}+9x-5 as \left(2x^{2}-x\right)+\left(10x-5\right).

$$\left(2x^{2}-x\right)+\left(10x-5\right)$$

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

$$x\left(2x-1\right)+5\left(2x-1\right)$$

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

$$\left(2x-1\right)\left(x+5\right)$$

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.

$$2x^{2}+9x-5=0$$

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.

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

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-9±\sqrt{81-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$x=\frac{-9±\sqrt{81+40}}{2\times 2}$$

Add 81 to 40.

$$x=\frac{-9±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$x=\frac{-9±11}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-9±11}{4}$$

Now solve the equation x=\frac{-9±11}{4} when ± is plus. Add -9 to 11.

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

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{-9±11}{4} when ± is minus. Subtract 11 from -9.

$$x=-\frac{20}{4}$$

Divide -20 by 4.

$$x=-5$$

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 -5 for x_{2}.

$$2x^{2}+9x-5=2\left(x-\frac{1}{2}\right)\left(x-\left(-5\right)\right)$$

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

$$2x^{2}+9x-5=2\left(x-\frac{1}{2}\right)\left(x+5\right)$$

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

$$2x^{2}+9x-5=2\times \frac{2x-1}{2}\left(x+5\right)$$

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

$$2x^{2}+9x-5=\left(2x-1\right)\left(x+5\right)$$

Evaluate

$$\left(2x-1\right)\left(x+5\right)$$