Solve a^3-4a^2-3a+18

$$a ^ { 3 } - 4 a ^ { 2 } - 3 a + 18$$

Factor

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

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 18 and q divides the leading coefficient 1. One such root is 3. Factor the polynomial by dividing it by a-3.

$$\left(a-3\right)\left(a^{2}-a-6\right)$$

Consider a^{2}-a-6. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-6. To find p and q, set up a system to be solved.

$$p+q=-1 pq=1\left(-6\right)=-6$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$p=-3 q=2$$

Rewrite a^{2}-a-6 as \left(a^{2}-3a\right)+\left(2a-6\right).

$$\left(a^{2}-3a\right)+\left(2a-6\right)$$

Factor out a in the first and 2 in the second group.

$$a\left(a-3\right)+2\left(a-3\right)$$

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

$$\left(a-3\right)\left(a+2\right)$$

Rewrite the complete factored expression.

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

Evaluate

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