Solve t^3-4t^2+3t

$$t ^ { 3 } - 4 t ^ { 2 } + 3 t$$

Factor

$$t\left(t-3\right)\left(t-1\right)$$

Solution Steps

Factor out t.

$$t\left(t^{2}-4t+3\right)$$

Consider t^{2}-4t+3. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt+3. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

Rewrite t^{2}-4t+3 as \left(t^{2}-3t\right)+\left(-t+3\right).

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

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

$$t\left(t-3\right)-\left(t-3\right)$$

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

$$\left(t-3\right)\left(t-1\right)$$

Rewrite the complete factored expression.

$$t\left(t-3\right)\left(t-1\right)$$

Evaluate

$$t\left(t-3\right)\left(t-1\right)$$