Solve (a)^2-5(a)+6

$$( a ) ^ { 2 } - 5 ( a ) + 6$$

Factor

$$\left(a-3\right)\left(a-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+6. To find p and q, set up a system to be solved.

$$p+q=-5 pq=1\times 6=6$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$p=-3 q=-2$$

Rewrite a^{2}-5a+6 as \left(a^{2}-3a\right)+\left(-2a+6\right).

$$\left(a^{2}-3a\right)+\left(-2a+6\right)$$

Factor out a in the first and -2 in the second group.

$$a\left(a-3\right)-2\left(a-3\right)$$

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

$$\left(a-3\right)\left(a-2\right)$$

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.

$$a^{2}-5a+6=0$$

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.

$$a=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6}}{2}$$

Square -5.

$$a=\frac{-\left(-5\right)±\sqrt{25-4\times 6}}{2}$$

Multiply -4 times 6.

$$a=\frac{-\left(-5\right)±\sqrt{25-24}}{2}$$

Add 25 to -24.

$$a=\frac{-\left(-5\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$a=\frac{-\left(-5\right)±1}{2}$$

The opposite of -5 is 5.

$$a=\frac{5±1}{2}$$

Now solve the equation a=\frac{5±1}{2} when ± is plus. Add 5 to 1.

$$a=\frac{6}{2}$$

Divide 6 by 2.

$$a=3$$

Now solve the equation a=\frac{5±1}{2} when ± is minus. Subtract 1 from 5.

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

Divide 4 by 2.

$$a=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and 2 for x_{2}.

$$a^{2}-5a+6=\left(a-3\right)\left(a-2\right)$$

Evaluate

$$\left(a-3\right)\left(a-2\right)$$