Solve 16j^4-40j^3+32j^5-20j^2

$$16 j ^ { 4 } - 40 j ^ { 3 } + 32 j ^ { 5 } - 20 j ^ { 2 }$$

Factor

$$4\left(2j+1\right)j^{2}\left(4j^{2}-5\right)$$

Solution Steps

Factor out 4.

$$4\left(4j^{4}-10j^{3}+8j^{5}-5j^{2}\right)$$

Consider 4j^{4}-10j^{3}+8j^{5}-5j^{2}. Factor out j^{2}.

$$j^{2}\left(4j^{2}-10j+8j^{3}-5\right)$$

Consider 4j^{2}-10j+8j^{3}-5. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -5 and q divides the leading coefficient 8. One such root is -\frac{1}{2}. Factor the polynomial by dividing it by 2j+1.

$$\left(2j+1\right)\left(4j^{2}-5\right)$$

Rewrite the complete factored expression. Polynomial 4j^{2}-5 is not factored since it does not have any rational roots.

$$4j^{2}\left(2j+1\right)\left(4j^{2}-5\right)$$

Evaluate

$$4\left(2j+1\right)j^{2}\left(4j^{2}-5\right)$$