Solve 6-5n+n^2

$$6 - 5 n + n ^ { 2 }$$

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$n^{2}-5n+6$$

Factor the expression by grouping. First, the expression needs to be rewritten as n^{2}+an+bn+6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\times 6=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b 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.

$$a=-3 b=-2$$

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

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

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

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

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

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

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.

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

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.

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

Square -5.

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

Multiply -4 times 6.

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

Add 25 to -24.

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

Take the square root of 1.

$$n=\frac{-\left(-5\right)±1}{2}$$

The opposite of -5 is 5.

$$n=\frac{5±1}{2}$$

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

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

Divide 6 by 2.

$$n=3$$

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

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

Divide 4 by 2.

$$n=2$$

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}.

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

Evaluate

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