Solve a^5+7a^4-18a^3=

$$a ^ { 5 } + 7 a ^ { 4 } - 18 a ^ { 3 } =$$

Factor

$$\left(a-2\right)\left(a+9\right)a^{3}$$

Solution Steps

Factor out a^{3}.

$$a^{3}\left(a^{2}+7a-18\right)$$

Consider a^{2}+7a-18. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-18. To find p and q, set up a system to be solved.

$$p+q=7 pq=1\left(-18\right)=-18$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 7.

$$p=-2 q=9$$

Rewrite a^{2}+7a-18 as \left(a^{2}-2a\right)+\left(9a-18\right).

$$\left(a^{2}-2a\right)+\left(9a-18\right)$$

Factor out a in the first and 9 in the second group.

$$a\left(a-2\right)+9\left(a-2\right)$$

Factor out common term a-2 by using distributive property.

$$\left(a-2\right)\left(a+9\right)$$

Rewrite the complete factored expression.

$$a^{3}\left(a-2\right)\left(a+9\right)$$

Evaluate

$$\left(a-2\right)\left(a+9\right)a^{3}$$