Solve 9b^6-4b^4-225b^2+100

$$9 b ^ { 6 } - 4 b ^ { 4 } - 225 b ^ { 2 } + 100$$

Factor

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

Solution Steps

Do the grouping 9b^{6}-4b^{4}-225b^{2}+100=\left(9b^{6}-4b^{4}\right)+\left(-225b^{2}+100\right), and factor out b^{4} in the first and -25 in the second group.

$$b^{4}\left(9b^{2}-4\right)-25\left(9b^{2}-4\right)$$

Factor out common term 9b^{2}-4 by using distributive property.

$$\left(9b^{2}-4\right)\left(b^{4}-25\right)$$

Consider 9b^{2}-4. Rewrite 9b^{2}-4 as \left(3b\right)^{2}-2^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(3b-2\right)\left(3b+2\right)$$

Consider b^{4}-25. Rewrite b^{4}-25 as \left(b^{2}\right)^{2}-5^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: b^{2}-5,b^{2}+5.

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

Evaluate

$$9b^{6}-4b^{4}-225b^{2}+100$$