| So how do you factor something | ||
| if you don't have a perfect square or cube? | ||
| How would we find the factors? | ||
| Would we take a guess? | ||
| Yep! That's one way to do it ... | ||
| Let's say we have the expression: | ||
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X3 + 4X2 + X - 6 |
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| Does this factor? | ||
| Who knows! | ||
| Let's guess that (X - 1) might be a factor. | ||
| Swell, how can you tell for sure? | ||
| Well, how would you check to see if 3 was a factor of 114? | ||
| You would divide 114 by 3 and see if it came out evenly. | ||
| That is, see if it divides with no remainder. | ||
| If it does, then 3 is a factor of 114. | ||
| Let's see ... | ||
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| Great 114 = 3 x 38. | ||
| How does that help us? | ||
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| We can use the same procedure to see if | ||
| X - 1 is a factor of X 3 + 4X 2 + X - 6. | ||
| We just divide the expression by X - 1. | ||
| SAY WHAT??? | ||
| It looks a little goofy, but trust me, it works. | ||
| First set up the problem ... | ||
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| The highest power of X in the whole expression is X 3. | ||
| The highest power of X in the thing we're dividing by is X. | ||
| X 3 divided by X is X 2. | ||
| That means, if this thing is going to work, | ||
| the first term we get from the division will be X 2. | ||
| Write it in ... | ||
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| See that we line it up over the 4X 2, | ||
| because it's the same power of X. | ||
| Now we multiply the term on top (the X 2) | ||
| by the thing we're dividing by (the X-1) | ||
| write the product under the expression | ||
| and subtract it from the expression. | ||
| This is exactly what we do in any division problem. | ||
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| So what's left starts with 5X 2. | ||
| If (X - 1) divides into this expression, | ||
| it will go 5X times, since the X from the (X - 1) | ||
| times 5X equals 5X 2. | ||
| THIS IS THE HEART OF THIS WHOLE | ||
| EQUATION DIVISION THING | ||
| Before you're done with this page, you need to understand | ||
| why the next term we write on top is 5X. | ||
| If you do, the rest is details. | ||
| If you don't, you don't know this concept. | ||
| Here's the next step ... | ||
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| Hey, this looks like it might work. | ||
| What's left starts with 6X. | ||
| If (X - 1) is going to divide into this, it will go 6 times | ||
| since 6 x X = 6X. | ||
| Let's see what we're left with. | ||
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| Since we have a remainder of zero, | ||
| we know that (X - 1) times (X 2 + 5X + 6) equals X 3 + 4X 2 + X - 6. | ||
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(X - 1) (X2 + 5X + 6) = X3 + 4X2 + X - 6 |
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| and if we look close, we might even see that | ||
| (X 2 + 5X + 6) factors to (X + 3)(X + 2). | ||
| That means we can say ... | ||
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(X - 1) (X + 3) (X + 2) = X3 + 4X2 + X - 6 |
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| So swell, fine, wonderful, this actually works. | ||
| But it has one big problem. | ||
| We had to guess the factor correctly to make this work. | ||
| If we had guessed, say, (X + 1) which is NOT a factor, | ||
| the division would not have come out evenly. | ||
| We would have to keep on guessing at factors | ||
| until we got lucky and guessed one right. | ||
| TRUE! | ||
| The thing is, that before computers came along | ||
| this was about the only way we had short of calculus | ||
| to find these factors. | ||
| Now a fancy calculator can check loads of factor guesses per second. | ||
| Unfortunately, you still need to learn this method. | ||
| Why? | ||
| Well, partly because lots of the time | ||
| you aren't looking for all of the factors of some big expression | ||
| you just want to know if some special thing is a factor | ||
| and that's easy to check. | ||
| Just divide and see if the remainder is zero. | ||
| Example: | ||
| Is 2X + 3 a factor of 10X 5 + 9X 4 - X 3 + 14X 2 - 17X - 12 | ||
| Another way to ask this is: Does 2X + 3 divide evenly into | ||
| 10X 5 + 9X 4 - X 3 + 14X 2 - 17X - 12 (without a remainder)? | ||
| Let's see ... | ||
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| 2X + 3 can't be divided evenly into 18, so we're stopped. | ||
| This one didn't work out evenly. | ||
| We have a remainder of 18. | ||
| That means 2X + 3 is not a factor. | ||
| If we want to get really technical, | ||
| we could use this remainder (18) | ||
| and the thing that we were dividing by (2X + 3) | ||
| to create a funny last term of our division | ||
| and have it work out evenly. | ||
| We could call the last term: | ||
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| Then when we multiply this by (2X + 3) we would get the 18 | ||
| that would make things work out evenly. | ||
| That would give is: | ||
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| While this little trick let us finish the problem, | ||
| most of the time we only want to call things factors | ||
| if they work out cleanly without the tricky fraction term at the end. | ||
|
copyright 2005 Bruce Kirkpatrick |
