Solve 16x-16+6x^2-12-6-4x^2+6x+10

$$16x-16+6 { x }^{ 2 } -12-6-4 { x }^{ 2 } +6x+10$$

Evaluate

$$2\left(x-1\right)\left(x+12\right)$$

Solution Steps

Subtract 12 from -16 to get -28.

$$16x-28+6x^{2}-6-4x^{2}+6x+10$$

Subtract 6 from -28 to get -34.

$$16x-34+6x^{2}-4x^{2}+6x+10$$

Combine 6x^{2} and -4x^{2} to get 2x^{2}.

$$16x-34+2x^{2}+6x+10$$

Combine 16x and 6x to get 22x.

$$22x-34+2x^{2}+10$$

Add -34 and 10 to get -24.

$$22x-24+2x^{2}$$

Factor

$$2\left(x-1\right)\left(x+12\right)$$

Solution Steps

Factor out 2.

$$2\left(11x-12+x^{2}\right)$$

Consider 8x-8+3x^{2}-6-3-2x^{2}+3x+5. Multiply and combine like terms.

$$x^{2}+11x-12$$

Consider x^{2}+11x-12. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=11 ab=1\left(-12\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 11.

$$a=-1 b=12$$

Rewrite x^{2}+11x-12 as \left(x^{2}-x\right)+\left(12x-12\right).

$$\left(x^{2}-x\right)+\left(12x-12\right)$$

Factor out x in the first and 12 in the second group.

$$x\left(x-1\right)+12\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+12\right)$$

Rewrite the complete factored expression.

$$2\left(x-1\right)\left(x+12\right)$$