Solve a^5-7a^4+12a^3

$$a ^ { 5 } - 7 a ^ { 4 } + 12 a ^ { 3 }$$

Factor

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

Solution Steps

Factor out a^{3}.

$$a^{3}\left(a^{2}-7a+12\right)$$

Consider a^{2}-7a+12. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+12. To find p and q, set up a system to be solved.

$$p+q=-7 pq=1\times 12=12$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$p=-4 q=-3$$

Rewrite a^{2}-7a+12 as \left(a^{2}-4a\right)+\left(-3a+12\right).

$$\left(a^{2}-4a\right)+\left(-3a+12\right)$$

Factor out a in the first and -3 in the second group.

$$a\left(a-4\right)-3\left(a-4\right)$$

Factor out common term a-4 by using distributive property.

$$\left(a-4\right)\left(a-3\right)$$

Rewrite the complete factored expression.

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

Evaluate

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