Solve s^2-14s+33

$${ s }^{ 2 } -14s+33$$

Factor

$$\left(s-11\right)\left(s-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as s^{2}+as+bs+33. To find a and b, set up a system to be solved.

$$a+b=-14 ab=1\times 33=33$$

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 33.

$$-1,-33 -3,-11$$

Calculate the sum for each pair.

$$-1-33=-34 -3-11=-14$$

The solution is the pair that gives sum -14.

$$a=-11 b=-3$$

Rewrite s^{2}-14s+33 as \left(s^{2}-11s\right)+\left(-3s+33\right).

$$\left(s^{2}-11s\right)+\left(-3s+33\right)$$

Factor out s in the first and -3 in the second group.

$$s\left(s-11\right)-3\left(s-11\right)$$

Factor out common term s-11 by using distributive property.

$$\left(s-11\right)\left(s-3\right)$$

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.

$$s^{2}-14s+33=0$$

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.

$$s=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 33}}{2}$$

Square -14.

$$s=\frac{-\left(-14\right)±\sqrt{196-4\times 33}}{2}$$

Multiply -4 times 33.

$$s=\frac{-\left(-14\right)±\sqrt{196-132}}{2}$$

Add 196 to -132.

$$s=\frac{-\left(-14\right)±\sqrt{64}}{2}$$

Take the square root of 64.

$$s=\frac{-\left(-14\right)±8}{2}$$

The opposite of -14 is 14.

$$s=\frac{14±8}{2}$$

Now solve the equation s=\frac{14±8}{2} when ± is plus. Add 14 to 8.

$$s=\frac{22}{2}$$

Divide 22 by 2.

$$s=11$$

Now solve the equation s=\frac{14±8}{2} when ± is minus. Subtract 8 from 14.

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

Divide 6 by 2.

$$s=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 11 for x_{1} and 3 for x_{2}.

$$s^{2}-14s+33=\left(s-11\right)\left(s-3\right)$$

Evaluate

$$\left(s-11\right)\left(s-3\right)$$