Solve t^4+9t^3+8t^2

$$t ^ { 4 } + 9 t ^ { 3 } + 8 t ^ { 2 }$$

Factor

$$\left(t+1\right)\left(t+8\right)t^{2}$$

Solution Steps

Factor out t^{2}.

$$t^{2}\left(t^{2}+9t+8\right)$$

Consider t^{2}+9t+8. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt+8. To find a and b, set up a system to be solved.

$$a+b=9 ab=1\times 8=8$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 9.

$$a=1 b=8$$

Rewrite t^{2}+9t+8 as \left(t^{2}+t\right)+\left(8t+8\right).

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

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

$$t\left(t+1\right)+8\left(t+1\right)$$

Factor out common term t+1 by using distributive property.

$$\left(t+1\right)\left(t+8\right)$$

Rewrite the complete factored expression.

$$t^{2}\left(t+1\right)\left(t+8\right)$$

Evaluate

$$\left(t+1\right)\left(t+8\right)t^{2}$$