Solve t^2-3t+2

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

Factor

$$\left(t-2\right)\left(t-1\right)$$

Solution Steps

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

$$a+b=-3 ab=1\times 2=2$$

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.

$$a=-2 b=-1$$

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

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

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

$$t\left(t-2\right)-\left(t-2\right)$$

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

$$\left(t-2\right)\left(t-1\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$t^{2}-3t+2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2}}{2}$$

Square -3.

$$t=\frac{-\left(-3\right)±\sqrt{9-4\times 2}}{2}$$

Multiply -4 times 2.

$$t=\frac{-\left(-3\right)±\sqrt{9-8}}{2}$$

Add 9 to -8.

$$t=\frac{-\left(-3\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$t=\frac{-\left(-3\right)±1}{2}$$

The opposite of -3 is 3.

$$t=\frac{3±1}{2}$$

Now solve the equation t=\frac{3±1}{2} when ± is plus. Add 3 to 1.

$$t=\frac{4}{2}$$

Divide 4 by 2.

$$t=2$$

Now solve the equation t=\frac{3±1}{2} when ± is minus. Subtract 1 from 3.

$$t=\frac{2}{2}$$

Divide 2 by 2.

$$t=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and 1 for x_{2}.

$$t^{2}-3t+2=\left(t-2\right)\left(t-1\right)$$

Evaluate

$$\left(t-2\right)\left(t-1\right)$$