Solve 2x^2-115x+1500

$$2 { x }^{ 2 } -115x+1500$$

Factor

$$\left(x-20\right)\left(2x-75\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+1500. To find a and b, set up a system to be solved.

$$a+b=-115 ab=2\times 1500=3000$$

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 3000.

$$-1,-3000 -2,-1500 -3,-1000 -4,-750 -5,-600 -6,-500 -8,-375 -10,-300 -12,-250 -15,-200 -20,-150 -24,-125 -25,-120 -30,-100 -40,-75 -50,-60$$

Calculate the sum for each pair.

$$-1-3000=-3001 -2-1500=-1502 -3-1000=-1003 -4-750=-754 -5-600=-605 -6-500=-506 -8-375=-383 -10-300=-310 -12-250=-262 -15-200=-215 -20-150=-170 -24-125=-149 -25-120=-145 -30-100=-130 -40-75=-115 -50-60=-110$$

The solution is the pair that gives sum -115.

$$a=-75 b=-40$$

Rewrite 2x^{2}-115x+1500 as \left(2x^{2}-75x\right)+\left(-40x+1500\right).

$$\left(2x^{2}-75x\right)+\left(-40x+1500\right)$$

Factor out x in the first and -20 in the second group.

$$x\left(2x-75\right)-20\left(2x-75\right)$$

Factor out common term 2x-75 by using distributive property.

$$\left(2x-75\right)\left(x-20\right)$$

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.

$$2x^{2}-115x+1500=0$$

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.

$$x=\frac{-\left(-115\right)±\sqrt{\left(-115\right)^{2}-4\times 2\times 1500}}{2\times 2}$$

Square -115.

$$x=\frac{-\left(-115\right)±\sqrt{13225-4\times 2\times 1500}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-115\right)±\sqrt{13225-8\times 1500}}{2\times 2}$$

Multiply -8 times 1500.

$$x=\frac{-\left(-115\right)±\sqrt{13225-12000}}{2\times 2}$$

Add 13225 to -12000.

$$x=\frac{-\left(-115\right)±\sqrt{1225}}{2\times 2}$$

Take the square root of 1225.

$$x=\frac{-\left(-115\right)±35}{2\times 2}$$

The opposite of -115 is 115.

$$x=\frac{115±35}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{115±35}{4}$$

Now solve the equation x=\frac{115±35}{4} when ± is plus. Add 115 to 35.

$$x=\frac{150}{4}$$

Reduce the fraction \frac{150}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{75}{2}$$

Now solve the equation x=\frac{115±35}{4} when ± is minus. Subtract 35 from 115.

$$x=\frac{80}{4}$$

Divide 80 by 4.

$$x=20$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{75}{2} for x_{1} and 20 for x_{2}.

$$2x^{2}-115x+1500=2\left(x-\frac{75}{2}\right)\left(x-20\right)$$

Subtract \frac{75}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2x^{2}-115x+1500=2\times \frac{2x-75}{2}\left(x-20\right)$$

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

$$2x^{2}-115x+1500=\left(2x-75\right)\left(x-20\right)$$

Evaluate

$$\left(x-20\right)\left(2x-75\right)$$