Solve 2x^2-8,x^3+8,x^2-2x-8

$$2 x ^ { 2 } - 8 , x ^ { 3 } + 8 , x ^ { 2 } - 2 x - 8$$

Least Common Multiple

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

Solution Steps

Factor the expressions that are not already factored.

$$-8+2x^{2}=2\left(x-2\right)\left(x+2\right) x^{3}+8=\left(x+2\right)\left(x^{2}-2x+4\right) -8-2x+x^{2}=\left(x-4\right)\left(x+2\right)$$

Identify all the factors and their highest power in all expressions. Multiply the highest powers of these factors to get the least common multiple.

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

Expand the expression.

$$2x^{5}-12x^{4}+16x^{3}+16x^{2}-96x+128$$

Evaluate

$$2\left(x^{2}-4\right),x^{3}+8,\left(x-4\right)\left(x+2\right)$$