Solve 3x^6-75x^4-48x^2+1200

$$3 x ^ { 6 } - 75 x ^ { 4 } - 48 x ^ { 2 } + 1200$$

Factor

$$3\left(x-5\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)\left(x^{2}+4\right)$$

Solution Steps

Factor out 3.

$$3\left(x^{6}-25x^{4}-16x^{2}+400\right)$$

Consider x^{6}-25x^{4}-16x^{2}+400. Do the grouping x^{6}-25x^{4}-16x^{2}+400=\left(x^{6}-25x^{4}\right)+\left(-16x^{2}+400\right), and factor out x^{4} in the first and -16 in the second group.

$$x^{4}\left(x^{2}-25\right)-16\left(x^{2}-25\right)$$

Factor out common term x^{2}-25 by using distributive property.

$$\left(x^{2}-25\right)\left(x^{4}-16\right)$$

Consider x^{2}-25. Rewrite x^{2}-25 as x^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Consider x^{4}-16. Rewrite x^{4}-16 as \left(x^{2}\right)^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Consider x^{2}-4. Rewrite x^{2}-4 as x^{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).

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

Rewrite the complete factored expression. Polynomial x^{2}+4 is not factored since it does not have any rational roots.

$$3\left(x-5\right)\left(x+5\right)\left(x-2\right)\left(x+2\right)\left(x^{2}+4\right)$$

Evaluate

$$3\left(x^{6}-25x^{4}-16x^{2}+400\right)$$