Solve a^2-3a+2

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

Factor

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

Solution Steps

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

$$p+q=-3 pq=1\times 2=2$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. The only such pair is the system solution.

$$p=-2 q=-1$$

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

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

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

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

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

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

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.

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

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.

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

Square -3.

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

Multiply -4 times 2.

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

Add 9 to -8.

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

Take the square root of 1.

$$a=\frac{-\left(-3\right)±1}{2}$$

The opposite of -3 is 3.

$$a=\frac{3±1}{2}$$

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

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

Divide 4 by 2.

$$a=2$$

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

$$a=\frac{2}{2}$$

Divide 2 by 2.

$$a=1$$

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}.

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

Evaluate

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