Solve 3x+8x-40-2x^2+10x

$$3x+8x-40-2 { x }^{ 2 } +10x$$

Evaluate

$$\left(8-x\right)\left(2x-5\right)$$

Solution Steps

Combine 3x and 8x to get 11x.

$$11x-40-2x^{2}+10x$$

Combine 11x and 10x to get 21x.

$$21x-40-2x^{2}$$

Factor

$$\left(8-x\right)\left(2x-5\right)$$

Solution Steps

Multiply and combine like terms.

$$-2x^{2}+21x-40$$

Factor the expression by grouping. First, the expression needs to be rewritten as -2x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=21 ab=-2\left(-40\right)=80$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 80.

$$1,80 2,40 4,20 5,16 8,10$$

Calculate the sum for each pair.

$$1+80=81 2+40=42 4+20=24 5+16=21 8+10=18$$

The solution is the pair that gives sum 21.

$$a=16 b=5$$

Rewrite -2x^{2}+21x-40 as \left(-2x^{2}+16x\right)+\left(5x-40\right).

$$\left(-2x^{2}+16x\right)+\left(5x-40\right)$$

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

$$2x\left(-x+8\right)-5\left(-x+8\right)$$

Factor out common term -x+8 by using distributive property.

$$\left(-x+8\right)\left(2x-5\right)$$