Solve 3a^5-12a^4+9a^3

$$3 a ^ { 5 } - 12 a ^ { 4 } + 9 a ^ { 3 }$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(a^{5}-4a^{4}+3a^{3}\right)$$

Consider a^{5}-4a^{4}+3a^{3}. Factor out a^{3}.

$$a^{3}\left(a^{2}-4a+3\right)$$

Consider a^{2}-4a+3. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+3. To find p and q, set up a system to be solved.

$$p+q=-4 pq=1\times 3=3$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. The only such pair is the system solution.

$$p=-3 q=-1$$

Rewrite a^{2}-4a+3 as \left(a^{2}-3a\right)+\left(-a+3\right).

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

Factor out a in the first and -1 in the second group.

$$a\left(a-3\right)-\left(a-3\right)$$

Factor out common term a-3 by using distributive property.

$$\left(a-3\right)\left(a-1\right)$$

Rewrite the complete factored expression.

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

Evaluate

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