Solve 2x^2-11x+15

$$2 { x }^{ 2 } -11x+15$$

Factor

$$\left(x-3\right)\left(2x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-11 ab=2\times 15=30$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -11.

$$a=-6 b=-5$$

Rewrite 2x^{2}-11x+15 as \left(2x^{2}-6x\right)+\left(-5x+15\right).

$$\left(2x^{2}-6x\right)+\left(-5x+15\right)$$

Factor out 2x in the first and -5 in the second group.

$$2x\left(x-3\right)-5\left(x-3\right)$$

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

$$\left(x-3\right)\left(2x-5\right)$$

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.

$$2x^{2}-11x+15=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\times 15}}{2\times 2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 2\times 15}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-11\right)±\sqrt{121-8\times 15}}{2\times 2}$$

Multiply -8 times 15.

$$x=\frac{-\left(-11\right)±\sqrt{121-120}}{2\times 2}$$

Add 121 to -120.

$$x=\frac{-\left(-11\right)±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$x=\frac{-\left(-11\right)±1}{2\times 2}$$

The opposite of -11 is 11.

$$x=\frac{11±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{11±1}{4}$$

Now solve the equation x=\frac{11±1}{4} when ± is plus. Add 11 to 1.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{11±1}{4} when ± is minus. Subtract 1 from 11.

$$x=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and \frac{5}{2} for x_{2}.

$$2x^{2}-11x+15=2\left(x-3\right)\left(x-\frac{5}{2}\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2x^{2}-11x+15=2\left(x-3\right)\times \frac{2x-5}{2}$$

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

$$2x^{2}-11x+15=\left(x-3\right)\left(2x-5\right)$$

Evaluate

$$\left(x-3\right)\left(2x-5\right)$$