# factoring polynomials

Let's review some factoring from the beginning algebra section.

You can use the distributive law to see that 3(4n + 5) = 3 * 4n + 3 * 5 = 12n + 15
and you can use FOIL to see that (n – 3)(n – 5) = n*n – 3*n  – n*5 + 3*5 = n^2 – 8n + 15.

It's like Jeopardy: I say, "n^2 – 8n + 15," and you respond (in the form of a question),
"What is (n – 3)(n – 5), Alex? I mean, Dan?"

Here's another way to look at the first problem:

12n + 15 = 3(4n) + 3(5) = 3(4n + 5) .

You go backwards, and this is called factoring out a common factor. Here's another:

10 h^2 + 15 hk = 5h * 2h + 5h * 3k = 5h (2h + 3k)

Example:  What about n^2 – 8n + 15 ?

This can be factored by finding two numbers whose sum is 8 and product is 15:

Let's use 3 and 5; it all works because of grouping (and cleverly choosing 3 and 5).

We get:   n^2 – 8n + 15

= n^2 – 3n – 5n + 15            now factor common factors in pairs:

= n(n  3) – 5(n – 3)            watch the signs!

= (n – 5)(n  3)                     the (n – 3) was the common factor!

Example:  Try to factor x^2 + 37x + 100.

Imagine x^2 + 37x + 100 = (x + m)(x + n).
We need two numbers m and n whose product is 100 and sum is 37.
Are you a list-maker?

```100 = (100)(1)... 100 + 1 = 101
100 = (50)(2) .... 50 + 2 = 52
100 = (25)(4) .... 25 + 4 = 29
100 = (20)(5) .... 20 + 5 = 25
100 = (10)(10)... 10 + 10 = 20```

It seems that 37 never comes up as a sum, so . . . x^2 + 37x + 100 doesn't factor.

But now do you see how you'd factor x^2 + 29x + 100 ?

Cleverly hidden answer: (x + 25)(x + 4).

Recall a linear equation is one that looks like ax + b = cx + d,  and our strategy was to get all x terms on the left, all constants on the right, then divide by the coefficient on x to solve.

A quadratic equation has an x^2 (x-squared) term; "quadrat" is Latin for square.

Note:  This web page was done originally on a Macintosh Quadra; the word is for the "040" processor chip in the computer.

The general quadratic equation looks like

a x^2 + b x + c = 0     where a ≠ 0.

If we want to find the x (or x's) that work, we might guess and substitute and hope we get lucky, or we might try one of these three methods:

• Factoring
• Completing the Square

Hey, let's solve the same equation three times, using each method!

Example (Factoring):

When the left-hand side factors, we can use the "zero product property" which says
"If a product is zero, one or more of the factors has to be zero."

x^2  8x + 15 = 0            our given equation

(x  3)(x  5) = 0             factor the left side

x – 3 = 0  or  x – 5 = 0    the "zero product property"

x = 3     or    x = 5             solving the two "little equations"

Example (Completing the square):

The idea is that: "stuff squared equals a number" is easy to solve, using square roots.
Notation:  I'll use -/(n) for square root of n, meaning the positive number whose square is n.

For example, -/49 = 7 because 7^2 = 49.

Also, -/ 2 is about 1.414, because 1.414^2 = 1.999396, which is close to 2. But you'll never hit 2 exactly by squaring a fraction (or terminating decimal), the square root of 2 is an "irrational number", meaning its decimal equivalent goes on forever (without a repeating block):

-/ 2 = 1.41421356237309504880168872420969807856967187537695 . . .

Now let's solve x^2 – 8x + 15 = 0 using the square root method. It will take four levels of trickiness to get there:

Level 1: Solve for x in the equation . . x^2 = 49.

x^2 = 49                      our given equation

-/(x^2) = -/49               take square root of both sides

x = + 7   or   –7            there are two numbers whose square is 49.

Level 2: Solve for x in the equation . . x^2 = 17.

x^2 = 17                             our given equation

-/(x^2) = -/17                      take square root of both sides

x = + -/17      or    – -/17     there are two numbers whose square is 17.
(we can leave the answer with square roots!)

Level 3: Solve for x in the equation . . (x + 6)^2 = 5.

(x + 6)^2 = 5                             our given equation

x + 6 = -/5   or   – -/5               take square root of both sides

x = – 6 + -/5   or   – 6 – -/5     subtracting 6 from both sides

Level 4:  Solve for x in the equation    x^2 – 8x + 15 = 0.

x^2 – 8x + 15 = 0                             what perfect square starts with x^2 – 8x ?

x^2 – 8x + 16 = 1

(x – 4)^2 = 1                                      now take square root of both sides

x – 4 = 1   or  1

x = 1 + 4 = 5    or   x = –1 + 4 = 3      add 4 to both sides.

So we ended up with the same solutions as before: x = 5  or  x = 3. Whew!

There's a formula for solving a general quadratic equation like a x^2 + b x + c = 0 ; it's called the quadratic formula
It gives the possible values for x in terms of a, b, and c.

You can derive it by completeing the square; the result is this:

If    a x^2 + b x + c = 0   then either

x = (b + -/(b^2  4ac)) / (2a)   or   x = (b – -/(b^2  4 ac)) / (2a)

The quantity D = (b^2 – 4 ac) is called the discriminant of the quadratic equation.

So x = (– b ± D) / (2a) .

• If D < 0 the equation has no real solutions, because we can't do a real square root of a negative number.
However, there are two complex solutions.
• If D = 0 the equation has one real (rational) solution x = - b / (2 a) because adding or subtracting 0 has no effect.
• If D > 0 then the equation has two real solutions
• If D is a perfect square (of a whole number or fraction) then there are two rational solutions for x.
This is when factoring will work. Trust me.

Now back to our original equation : x^2 – 8x + 15 = 0 . In the above notation, we have a = 1, b = -8, and c = 15.
Using the quadratic formula gives us

x = ( –b ± -/(b^2 – 4ac)) / (2a)

= (–(–8) ± -/((–8)^2 – 4(1)(15))) / (2(1))

= (8 ± -/(64 – 60)) / 2 = (8 ± -/4) / 2

= (8 + 2) / 2 or (8 – 2) / 2 ; so again x = 5 or x = 3   ;-}

But the power of the quadratic formula is its ability to solve equations that don't factor.

Example: The Golden Ratio: phi = 1.618... is one of the roots of x^2  x  1 = 0 ;
use the quadratic formula and see that the roots are x = (1 ± -/5) / 2 .
This is closely related to the Fibonacci Sequence.

How can we draw a picture of y = x^2 ? Like any graph, we put a buncha dots at (x, x^2), and notice the pattern.

We can make a table and go from there. For example, if x = –2 then y = (–2)^2 = 4.

Locate the list of points on the graph.

•  x    x^2
• –2    4
• –1     1
•  0     0
•  1      1
•  2     4
•  3     9