Solve 2b^5-100b^3+98b

$$2 b ^ { 5 } - 100 b ^ { 3 } + 98 b$$

Factor

$$2b\left(b-7\right)\left(b-1\right)\left(b+1\right)\left(b+7\right)$$

Solution Steps

Factor out 2.

$$2\left(b^{5}-50b^{3}+49b\right)$$

Consider b^{5}-50b^{3}+49b. Factor out b.

$$b\left(b^{4}-50b^{2}+49\right)$$

Consider b^{4}-50b^{2}+49. Find one factor of the form b^{k}+m, where b^{k} divides the monomial with the highest power b^{4} and m divides the constant factor 49. One such factor is b^{2}-49. Factor the polynomial by dividing it by this factor.

$$\left(b^{2}-49\right)\left(b^{2}-1\right)$$

Consider b^{2}-49. Rewrite b^{2}-49 as b^{2}-7^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(b-7\right)\left(b+7\right)$$

Consider b^{2}-1. Rewrite b^{2}-1 as b^{2}-1^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(b-1\right)\left(b+1\right)$$

Rewrite the complete factored expression.

$$2b\left(b-7\right)\left(b+7\right)\left(b-1\right)\left(b+1\right)$$

Evaluate

$$2b\left(b^{4}-50b^{2}+49\right)$$