Solve 2x^2-15x+25

$$2 { x }^{ 2 } -15x+25$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=-15 ab=2\times 25=50$$

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 50.

$$-1,-50 -2,-25 -5,-10$$

Calculate the sum for each pair.

$$-1-50=-51 -2-25=-27 -5-10=-15$$

The solution is the pair that gives sum -15.

$$a=-10 b=-5$$

Rewrite 2x^{2}-15x+25 as \left(2x^{2}-10x\right)+\left(-5x+25\right).

$$\left(2x^{2}-10x\right)+\left(-5x+25\right)$$

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

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

Factor out common term x-5 by using distributive property.

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

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.

$$2x^{2}-15x+25=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 2\times 25}}{2\times 2}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 2\times 25}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-15\right)±\sqrt{225-8\times 25}}{2\times 2}$$

Multiply -8 times 25.

$$x=\frac{-\left(-15\right)±\sqrt{225-200}}{2\times 2}$$

Add 225 to -200.

$$x=\frac{-\left(-15\right)±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-\left(-15\right)±5}{2\times 2}$$

The opposite of -15 is 15.

$$x=\frac{15±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{15±5}{4}$$

Now solve the equation x=\frac{15±5}{4} when ± is plus. Add 15 to 5.

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

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{15±5}{4} when ± is minus. Subtract 5 from 15.

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

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 5 for x_{1} and \frac{5}{2} for x_{2}.

$$2x^{2}-15x+25=2\left(x-5\right)\left(x-\frac{5}{2}\right)$$

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

$$2x^{2}-15x+25=2\left(x-5\right)\times \frac{2x-5}{2}$$

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

$$2x^{2}-15x+25=\left(x-5\right)\left(2x-5\right)$$

Evaluate

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