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Factoring Polynomials of Degree 3

```
Date: 3/17/96 at 15:50:7
From: Ian Rounthwaite
Subject: factoring

1. Factor completely (x+2)3 (to the third) + (x+2)2 (to the second)

2. Factor completely 18x(squared) + 9x - 2

Thank you,
Barb R.
```

```
Date: 3/17/96 at 19:8:42
From: Doctor Jodi
Subject: Re: factoring

Hi there Barb!

Well, in your first problem the first step is to multiply everything
out.(By the way, on computers, we usually write exponents like this:

So, if we want to find out what (x+2)^3 + (x+2)^2  is,

first we multiply (x+2)(x+2) Do you know how to do this?

we can also write (x+2)(x+2) as x(x+2) + 2(x+2)
which equals                    x^2+2x + 2x+4

We can simplify this to x^2 + 4x + 4, since we have two terms (2x
and 2x) which are both multiplied by x.

Does this make sense so far?

Well, know we know one half of the problem:

(x+2)^3        +      (x+2)^2 =

(x+2)^3        +      x^2 + 4x + 4 (from our multiplication).

Now, to find out what (x+2)^3 is, we need to multiply

(x+2)(x+2)(x+2)

Well, we already know that (x+2)(x+2) = x^2+4x+4 (from our
multiplication above). Now we need to multiply that by another
(x+2):

(x+2) (x^2 + 4x +4)

which we can also write as

x(x^2 + 4x + 4) + 2 (x^2 + 4x + 4) (Do you understand why?)

Now, multiplying, we get x^3 + 4x^2 + 4x + 2x^2 + 8x + 8

Now, if we add like terms (like 4x^2 and 2x^2, etc)

We get x^3 + 6X^2 + 12 x + 8 = (x+2)^3

Okay?

But we need to put this answer back into our original equation where

(x+2)^3 was.

(X+2)^3 + x^2 + 4x + 4

We now have x^3 + 6x^2 + 12x + 8 + x^2 + 4x + 4
^^^^^^^^^^^^^^^^^^^^   ^^^^^^^^^^^^

from (x+2)^3      from (x+2)^2

Okay. At this point, we're going to add like terms again.

So after we do this, we have

x^3 + 7x^2 + 16x + 12   right?

NOW comes the tricky part: the factoring.

There are two methods:

trial and error (which means, you try a couple of likely
multiplications)

and

division.

I tried trial and error for a while but I didn't come up with an

So then I tried division.

Maybe you haven't seen division before in this context:  it's pretty
similar to division with plain old numbers, but much cooler.

It looks something like this.

I think that our polynomial (x^3 + 7x^2 + 16x + 12) is going to be
divisible by x+2 (since we multiplied out some powers of x+2 and

So I try this

___________________________
x+2 |x^3 + 7x^2 + 16x +12

first,

I look at x^3 + 7x^2.  How many times will x+2 go into x^3 + 7x^2?

I guess that it's around x^2 times, since x+2 times x^2 = x^3 + 2x^2

So now I write that down and subtract:

x^2
_________________________
x+2 |x^3 + 7x^2 + 16x +12
-(x^3 + 2x^2)
___________
5x^2

Next I bring down the next term:

x^2
___________________________
x+2 |x^3 + 7x^2 + 16x +12
-(x^3 + 2x^2)
_____________
5x^2 + 16x

and guess again:

This time, my guess is 5x. Can you see why?

x^2 + 5x
___________________________
x + 2 |x^3 + 7x^2 + 16x +12
-(x^3 + 2x^2)
__________
5x^2 + 16x
-(5x^2 + 10x)
_________
6x

Next, I'll bring down the 12 and finish up with a guess of 6.

x^2 + 5x + 6
_______________________________
x+2|x^3 + 7x^2 + 16x +12
-(x^3 + 2x^2)
_____________
5x^2 + 16x
- (5x^2 + 10x)
______________
6x         +12
-(6x +12)
_________
0

Whew! Looks like it worked... We can check by multiplying

(x+2) (x^2 + 5x +6)

Now we've factored x^3 + 7X^2 + 16x +12

To FULLY factor, we'll need to see if either x+2 or x^2 + 5x +6
have factors.

Intuition tells me that x^2 + 5x + 6 has some factors. Think you
can handle this part on your own?

Do you think that you'll be okay with the second problem now?

If you need more help, write us back!

-Doctor Jodi,  The Math Forum
```

```
Date: 3/21/96 at 13:58:53
From: "Ian Rounthwaite"
Date: Thu, 21 Mar 1996 11:54:28 MST
Subject: help

Hi - my question ((x+2)^3 + (x+2)^2 - you took me as far as
x^2+5x-6 and said to factor more.   So I think the answer is
x(x+5)-6.  I understood your answer for the rest I think.  It was

Thank you.  Also, question 2 was 18x^2 +9x-2. (factor completely like
the other one). Is the answer 9x(2x+1)-2?

Thanks, Barb R.
```

```
Date: 3/21/96 at 23:53:57
From: Doctor Patrick
Subject: Re: help

Hi  there.  Sorry, that's not quite right.  When you factor a problem,
you have to factor the entire polynomial, not just parts of it, just
like when you find the factors of a whole number.  If you were going
to factor 6, you would say it was 3*2, not 2*2 +2, even though they
both equal 6.

Likewise, when you factor you need to break it into two parts that
when multiplied together will give you the whole polynomial back.
I'll show you how to do this part, and then let you do the second on

One thing before we start - when you wrote back to us you were
using x^2+5x-6 - instead of x^2+5x + 6 which was the correct
polynomial. I'll use that to demonstrate the problem.

Let me give you the answer first, so you know what we are looking
for: (x+3)*(x+2). Do you know how to multiply these? The way you
do it is to multiply both of the numbers in the first half by both of
the numbers of the second half and add the results when you multiply
this you get (x*x) - (2x) +(3x) +(6), which gives us our x^2+5x-6
when we add like terms and multiply the x's.  .

Now, on to how to get the answer.  To get the x^2 we will half to
multiply an x times an x, so we can safely say that the factors will
look like:

(x [+ or -] some number) * (x [+ or -] some other  number)

Further, we know that the second numbers must be factors of 6 in
order to give us the 6 in the final equation.  There factors will add
up to the 5 which is in the middle term of the polynomial.  Since the
6 is positive, it will have to have either both positive, or both
negative, factors.  And since the 5 is positive, and can only be made
by adding at least one positive number (either a positive and a
negative, or a positive and a lesser negative) then the factors can only
be positive.  So now our factors are:

(x+ some factor of 6) * (x+ another factor of 6).

Since the only factor of six that add up to 5 are 3 and 2 we have for

Also - don't forget about the other (x+2) left over from before -
that still part of the number too, making the final answer
(x+2)*(x+2)*(x+3), or (x+2)^2 * (x+3).  Does that make sense
to you?

One last thing  There is an easier way to factor the first problem
than our last answer gave.  Whenever I have to factor, the first thing
I look for is a common factor of all of the terms so I can divide it
out and simplify the expression.  What do you think it might be in
this problem?

Let's see - (X+2)^3 could be rewritten as (X+2)(X+2)^2.  Can you see why?

If you're not sure, try rewriting (X+2)(X+2)^2 without using the
expound, and then do the same for (X+2)^3 and compare what you
get.

This makes our expression (X+2)(X+2)^2 +(X+2)^2.  The (X+2)^2
is common to both terms, so we can factor it out, making our
expression (X+2)^2 ((X+2) +1).  We can simplify this further by
re-writing it as (X+2)^2 (X+2 +1) and then adding the like terms.

This gives us (X+2)^2 (X+3), which is the final answer

-Doctor Patrick,  The Math Forum
```

```
Date: 4/2/96 at 18:42:30
From: "Ian Rounthwaite"
Subject: help

Factor completely

18x^2+9x-2
Barb R.
Thanks.
```

```
Date: 4/16/96 at 19:27:35
From: Doctor Patrick
Subject: Re: help

Hello again!  To factor 18x^2+9x-2 we need to find two terms
which, when multiplied, will give us back our original equation.

Whenever I factor this kind of problem, I start by looking for
the possible factors for the number before the x^2 term (in this
case 18) and the number which is not multiplied by a varible
(2 in your problem). Then I look to see which combinations of
the two will give me the middle term (9x).

For example 18x^2 can be factored into 18x*1x, 9x*2x, 6x*3x, or
-18x*-1x, -9x*-x2, and -6x*-3x. -2 can only be factored to -2*1
or -1*2. We now have to find a way to combine the two parts in
order to make 9.  However we combine it, since the middle term
is positive we know that when we start multiplying the largest product
will have to be positive also.  Do you understand why?

(18x+2)(1x-1)=18x^2-17x-2.  That didn't work very well, did it?
We could try some other combinations with the 18, but I have the
feeling it won't work, since we only have the 2 to add to or subtract
from it.  I don't think that that will be enough, do you?

What we are left with at this point are the 9x*2x and 6x*3x
combinations. There really isn't a way to tell which will work so
a lot of what follows is guesswork and trial and error.  You have to
play with the different combinations to figure out which will work.
Often, you can rule out an entire set of possibilities by looking at
some of its results, as we did above.

It turns out that the 9x*2x combinations don't work either, but I'll
let you test those on your own in order to keep this answer from
getting too long.  All we have left are 6x*3x combinations to make
the 18x^2.  I'll let you work out these on your own - good luck!

Write us back if you need more help!

-Doctor Patrick,  The Math Forum

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Associated Topics:
Middle School Factoring Expressions

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