Algebra 1 Factoring Polynomials
Math-Prof HOME Algebra 1 Table of Contents Ask A Question PREV NEXT

Heavy Duty Factoring
Factoring Polynomials

 

 OK, this has all been great up to now,

 but where this stuff really gets used
 is when we have something like:
 

X2 + 12X + 32

 
 and we try to figure out what two "factors"
 someone might have multiplied together to get this.
 
 When we start working on one of these
 we don't know if we can find two factors
 but let's try.
 
 To wind up with an X 2 term, 
 we need to multiply X times X.
 
 That means that if we are going to be able to factor this thing,
 the factors will start out like this:
 

(X         )(X         )

 
 The next thing we do is see 
 that all of the terms are positive.
 
 That means if we CAN factor this
 all of the factors will be positive numbers.
 That means we can put in a bit more info ...
 

(X  +       )(X  +       )

 
 The last (and hardest) part is to figure out
 what number we need to finish off the factors.
 
 If any numbers will work,
 they will be numbers that multiplied together
 give us the last number in our original problem.
 That number was 32.
 
 So we line up all of the pairs of numbers 
 that multiplied together give us 32.
 They are:
 

1 32     2 16     4 8

 
 For the numbers to work, 
 they also have to add up to the number part 
 of the middle term (the 12).
 
 So let's check to see if any of them do ...
 

 1 + 32 = 33     2 + 16 = 18     4 + 8 = 12

 
 So 4 and 12 are our winners.
 The two factors we have built so far are both the same
 so it doesn't matter which number goes where.
 
 So:

X2 + 12X + 32 = (X + 4)(X + 8)

 
 Example:
 
 Let's try to factor 

 X2 - 4X - 32

 
 The first term is still X 2, so to get it
 the factors still need to start out with:
 

(X         )(X         )

 
 Since we have some negative numbers
 in the thing we are trying to factor
 we know that there has to be a negative
 in here somewhere.
 
 Since the 32 is negative, 
 we know that only one of the two numbers is negative.
 
 If both of the numbers had been negative,
 the 32 would have wound up being positive.
 Since both of the factors are the same so far
 the negative can go on either one.

 So we have ...

 (X  +       )(X  -       )

 
 Now we need to figure out what the numbers are.
 The term on the end is 32 again, 
 so the numbers that might work 
 are the same as before.
 (Don't worry about the minus sign yet)
 

1 x 32     2 x 16     4 x 8

 
 NOW, since we have one negative and one positive number
 we are looking for numbers where the difference
 is -4.
 That's pretty easy to spot.
 

 4 - 8 = -4

 
 So the numbers we need are 4 and -8.
 That makes the factors:
 

(X + 4)(X - 8)

 
 That makes the whole thing ...
 

X2 - 4X - 32 = (X + 4)(X - 8)

 
 Remember, there's no guarantee 
 that we will be able to factor these things.
 If we can't find a pair of numbers
 that multiply together to give us the last term
 and the sum or difference gives us the number part
 of the middle term
 we can't factor the expression.
 At least with the tricks we have so far ...
 
 Great, Let's try one more.
 
 Example:
 
 Find the factors of:
 

2X2 - 15X + 18

 
 OK, this one is a little trickier.
 We have 2X 2 now instead of X 2.
 To get 2X 2, we need to multiply 2X times X.
 That means our factors start out as:
 

(X         )(2X         )

 
 The sign on the middle term is negative
 so we know that there has to be at least one 
 negative number in the factors.
 
 The third term is a positive number.
 We know that since at least one of the number factors must be negative
 and the third term is positive
 that BOTH of the number factors must be negative.
 That means our factors so far are:
 

(X  -      )(2X  -      )

 
 Now we need to know what pairs of numbers
 can be multiplied together to get 18 ...
 

1 x 18     2 x 9     3 x 6

 
 Now we're ready for the middle term (addition) part.
 Up to now we've just tried adding or subtracting
 each of the multiplication pairs to find the middle term number.
 Here, the X factors are X and 2X.
 that means in this problem one of the numbers
 gets included TWICE when the middle term number gets built.
 
 OK, the number we are trying to get to is -15.
 One of the numbers will count twice.
 

 - 1 - 1 - 18 = - 20

- 1 - 18 - 18 = - 37
 - 2 - 2 - 9 = - 13
 - 2 - 9 - 9 = - 20
 - 3 - 3 - 6 = - 12
 - 3 - 6 - 6 = -18

  We have a winner!

 
 So we need to use the 3 and the 6.
 We also need to arrange them 
 so that the 6 gets multiplied times the 2.
 That means the 6 needs to be in the other factor
 than the one the 2X is in.
 
 So the factors are:
 

(X - 6)(2X - 3)

 
 Which makes the whole thing ...
 

 2X2 - 15X + 18 = (X - 6)(2X - 3)

 

   copyright 2005 Bruce Kirkpatrick

Math-Prof HOME Algebra 1 Table of Contents Ask A Question PREV NEXT