Solve 2g^3-g^2-8g+4

$$2 g ^ { 3 } - g ^ { 2 } - 8 g + 4$$

Factor

$$\left(g-2\right)\left(2g-1\right)\left(g+2\right)$$

Solution Steps

Do the grouping 2g^{3}-g^{2}-8g+4=\left(2g^{3}-g^{2}\right)+\left(-8g+4\right), and factor out g^{2} in the first and -4 in the second group.

$$g^{2}\left(2g-1\right)-4\left(2g-1\right)$$

Factor out common term 2g-1 by using distributive property.

$$\left(2g-1\right)\left(g^{2}-4\right)$$

Consider g^{2}-4. Rewrite g^{2}-4 as g^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(g-2\right)\left(g+2\right)$$

Rewrite the complete factored expression.

$$\left(g-2\right)\left(2g-1\right)\left(g+2\right)$$

Evaluate

$$\left(2g-1\right)\left(g^{2}-4\right)$$