Solve 5a^4+65a^3+200a^2

$$5 a ^ { 4 } + 65 a ^ { 3 } + 200 a ^ { 2 }$$

Factor

$$5\left(a+5\right)\left(a+8\right)a^{2}$$

Solution Steps

Factor out 5.

$$5\left(a^{4}+13a^{3}+40a^{2}\right)$$

Consider a^{4}+13a^{3}+40a^{2}. Factor out a^{2}.

$$a^{2}\left(a^{2}+13a+40\right)$$

Consider a^{2}+13a+40. Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+40. To find p and q, set up a system to be solved.

$$p+q=13 pq=1\times 40=40$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 40.

$$1,40 2,20 4,10 5,8$$

Calculate the sum for each pair.

$$1+40=41 2+20=22 4+10=14 5+8=13$$

The solution is the pair that gives sum 13.

$$p=5 q=8$$

Rewrite a^{2}+13a+40 as \left(a^{2}+5a\right)+\left(8a+40\right).

$$\left(a^{2}+5a\right)+\left(8a+40\right)$$

Factor out a in the first and 8 in the second group.

$$a\left(a+5\right)+8\left(a+5\right)$$

Factor out common term a+5 by using distributive property.

$$\left(a+5\right)\left(a+8\right)$$

Rewrite the complete factored expression.

$$5a^{2}\left(a+5\right)\left(a+8\right)$$

Evaluate

$$5\left(a+5\right)\left(a+8\right)a^{2}$$