Solve 40x-80+6x+12-3x^2+6x-6x+12

$$40x-80+6x+12-3 { x }^{ 2 } +6x-6x+12$$

Evaluate

$$\left(14-x\right)\left(3x-4\right)$$

Solution Steps

Combine 40x and 6x to get 46x.

$$46x-80+12-3x^{2}+6x-6x+12$$

Add -80 and 12 to get -68.

$$46x-68-3x^{2}+6x-6x+12$$

Combine 46x and 6x to get 52x.

$$52x-68-3x^{2}-6x+12$$

Combine 52x and -6x to get 46x.

$$46x-68-3x^{2}+12$$

Add -68 and 12 to get -56.

$$46x-56-3x^{2}$$

Factor

$$\left(14-x\right)\left(3x-4\right)$$

Solution Steps

Multiply and combine like terms.

$$-3x^{2}+46x-56$$

Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx-56. To find a and b, set up a system to be solved.

$$a+b=46 ab=-3\left(-56\right)=168$$

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 168.

$$1,168 2,84 3,56 4,42 6,28 7,24 8,21 12,14$$

Calculate the sum for each pair.

$$1+168=169 2+84=86 3+56=59 4+42=46 6+28=34 7+24=31 8+21=29 12+14=26$$

The solution is the pair that gives sum 46.

$$a=42 b=4$$

Rewrite -3x^{2}+46x-56 as \left(-3x^{2}+42x\right)+\left(4x-56\right).

$$\left(-3x^{2}+42x\right)+\left(4x-56\right)$$

Factor out 3x in the first and -4 in the second group.

$$3x\left(-x+14\right)-4\left(-x+14\right)$$

Factor out common term -x+14 by using distributive property.

$$\left(-x+14\right)\left(3x-4\right)$$