Solve 4x^5-25x^3+32x^2-200

$$4 x ^ { 5 } - 25 x ^ { 3 } + 32 x ^ { 2 } - 200$$

Factor

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

Solution Steps

Do the grouping 4x^{5}-25x^{3}+32x^{2}-200=\left(4x^{5}-25x^{3}\right)+\left(32x^{2}-200\right), and factor out x^{3} in the first and 8 in the second group.

$$x^{3}\left(4x^{2}-25\right)+8\left(4x^{2}-25\right)$$

Factor out common term 4x^{2}-25 by using distributive property.

$$\left(4x^{2}-25\right)\left(x^{3}+8\right)$$

Consider 4x^{2}-25. Rewrite 4x^{2}-25 as \left(2x\right)^{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).

$$\left(2x-5\right)\left(2x+5\right)$$

Consider x^{3}+8. Rewrite x^{3}+8 as x^{3}+2^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

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

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

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

Evaluate

$$4x^{5}-25x^{3}+32x^{2}-200$$