Solve (-t^2+3t)+(t-3)

$$( - t ^ { 2 } + 3 t ) + ( t - 3 )$$

Factor

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

Solution Steps

Multiply and combine like terms.

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

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.

$$a+b=4 ab=-\left(-3\right)=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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 -t^{2}+3t.

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

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

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

Evaluate

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

Solution Steps

Combine 3t and t to get 4t.

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