*dan's intermediate algebra lesson*

*dan's intermediate algebra lesson*

* ©1997-2016 - all rights reserved - dan bach @dansmath*

**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**.

But how can you start with the answer and find the factors?

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 =101100 = (50)(2) .... 50 + 2 =52100 = (25)(4) .... 25 + 4 =29100 = (20)(5) .... 20 + 5 =25100 = (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)*.

**Solving Quadratic Equations**

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****The Quadratic Formula**

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!

**Example (The Quadratic Formula):**

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.

**Graphing parabolas **(See also functions and graphs.)

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.

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