Solve a^{2}-2a^2+3a^4-4a^5+6a^5=

$$a ^ { 2 } - 2 a ^ { 2 } + 3 a ^ { 4 } - 4 a ^ { 5 } + 6 a ^ { 5 } =$$

Evaluate

$$\left(2a-1\right)\left(a\left(a+1\right)\right)^{2}$$

Solution Steps

Combine a^{2} and -2a^{2} to get -a^{2}.

$$-a^{2}+3a^{4}-4a^{5}+6a^{5}$$

Combine -4a^{5} and 6a^{5} to get 2a^{5}.

$$-a^{2}+3a^{4}+2a^{5}$$

Factor

$$\left(2a-1\right)a^{2}\left(a+1\right)^{2}$$

Solution Steps

Factor out a^{2}.

$$a^{2}\left(-1+3a^{2}+2a^{3}\right)$$

Consider 1-2+3a^{2}-4a^{3}+6a^{3}. Multiply and combine like terms.

$$2a^{3}+3a^{2}-1$$

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

$$\left(2a-1\right)\left(a^{2}+2a+1\right)$$

Consider a^{2}+2a+1. Use the perfect square formula, p^{2}+2pq+q^{2}=\left(p+q\right)^{2}, where p=a and q=1.

$$\left(a+1\right)^{2}$$

Rewrite the complete factored expression.

$$a^{2}\left(2a-1\right)\left(a+1\right)^{2}$$