Solve a^3-5a^2+4a

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

Factor

$$a\left(a-4\right)\left(a-1\right)$$

Solution Steps

Factor out a.

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

Consider a^{2}-5a+4. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+4. To find p and q, set up a system to be solved.

$$p+q=-5 pq=1\times 4=4$$

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 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$p=-4 q=-1$$

Rewrite a^{2}-5a+4 as \left(a^{2}-4a\right)+\left(-a+4\right).

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

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

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

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

$$\left(a-4\right)\left(a-1\right)$$

Rewrite the complete factored expression.

$$a\left(a-4\right)\left(a-1\right)$$

Evaluate

$$a\left(a-4\right)\left(a-1\right)$$