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Finding the Sum of Solutions of a Polynomial Equation

```Date: 03/11/2004 at 21:04:54
From: Zahra
Subject: x^3+2x^2-5x-6=0

What is the sum of the solutions of x^3 + 2x^2 - 5x - 6 = 0 given that
x = 2 is a solution?  I do not understand any part of this question.

```

```
Date: 03/12/2004 at 14:01:18
From: Doctor Ian
Subject: Re: x^3+2x^2-5x-6=0

Hi Zahra,

Let's look at a simpler example:

What is the sum of the solutions of x^2 - 5x + 6 = 0,
given that x = 2 is a solution?

If x = 2 is a solution, then

x^2 - 5x + 6 = (x - 2)(x +/- something)

So what we want to do is find the "something".  (Do you see why this
is true?  Write back if you don't, and we can talk about that.)

Now, how do we find it?  Well, suppose I tell you that

12 = 4 * something

How could you find the "something"?  You'd divide:

12 / 4 = something

= 3

Same thing here.  We want to divide x^2 - 5x + 6 by (x - 2).  The
process works just like long division with numbers, except you deal
with one term at a time instead of one digit at a time:

x
_____________
x - 2 ) x^2 - 5x + 6
x^2 - 2x          <- This is x * (x - 2)
--------
-3x          <- This is -5x - (-2x), or -5x + 2x

Now we bring the next term down, and keep going:

x - 3
_____________
x - 2 ) x^2 - 5x + 6
x^2 - 2x
--------
-3x + 6
-3x + 6
-------
0

There's no remainder, so

x^2 - 5x + 6 = (x - 2)(x - 3)

If you multiply out the right side, you'll see that this is, in fact,
correct.

To finish our problem, the two solutions to the equation are 2 and 3,
so their sum is 5.

Now, with a quadratic, this is overkill.  It's easier to use regular
factoring methods.  However, for higher-degree polynomials, it's a
useful technique.  If you have a cubic, and you know one of the roots
is x = R, you can divide by (x - R) to get a quadratic, which you can
then factor or solve normally.

Or if you have a quartic polynomial, and you can find a root, you can
divide to get a cubic; then if you can find another root, you can
divide to get a quadratic.  And so on.

Does it make more sense now?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 03/12/2004 at 19:48:56
From: Zahra
Subject: Thank you (x^3+2x^2-5x-6=0)

Dear Dr. Ian:

I just wanted to thank you for the great help.  It was actually the
first time I went on a math web site, and I loved it.  I like the way
you explained the question to me.  I am 14 years old and I'm in 8th
grade.  This was my first year going to the Mu alpha competitions in
Florida.  I love it, and I will be there every year!

Sincerely,
Zahra

```

```
Date: 03/14/2004 at 19:34:29
From: Zahra
Subject: x^3+2x^2-5x-6=0

Dear Dr. Ian -

I understand the whole process you explained.  But when I tried the
original question I asked you it doesn't work.  Here is what I did:

x^2 + 4x + 3
_________________
x - 2 )x^3 + 2x^2 - 5x - 6
x^3 - 2x^2
-------------
4x^2 - 5x - 6
4x^2 - 8x
----------
3x - 6
3x - 6
------
0

Now, I don't think my answer is right.  But I thought maybe I have to
factor the answer I got, so I did.  Here is what I got:

x^2 + 4x + 3
(x + 1)(x + 3)

But this can not be right because they say "given that x = 2".  Thanks
for the great help.
Zahra

```

```
Date: 03/14/2004 at 21:09:13
From: Doctor Ian
Subject: Re: x^3+2x^2-5x-6=0

Hi Zahra,

Actually, you've done all the steps perfectly!  The only thing that
remains is to see what the results are telling us.

Recall that the first division tells us that

x^3 + 2x^2 - 5x - 6 = (x - 2)(x^2 + 4x + 3)

And you factored the quadratic, which shows that

x^3 + 2x^2 - 5x - 6 = (x - 2)(x + 1)(x + 3)

Note that for the expression on the left to be equal to zero, the
expression on the right must also be equal to zero.  But when is that
true?  The only way to get a product of zero is for one or more of the
factors to be equal to zero.  (Do you see why?)

We've been told that x = 2 is a solution, and indeed, substituting 2
for x gives us a value of zero:

(2 - 2)(2 + 1)(2 + 3) = 0 * 3 * 5

= 0

But there are two other solutions as well.  One of them is x = -1,
because adding -1 to 1 will make the second term zero:

(-1 - 2)(-1 + 1)(-1 + 3) = -3 * 0 * 2

= 0

Can you see what the third solution has to be?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 03/15/2004 at 21:28:45
From: Zahra
Subject: x^3+2x^2-5x-6=0

Dear Dr. Ian:

Since (x + 1) gives a solution of -1 and (x - 2) gives 2, I would say
that the third solution from (x + 3) is -3.  Therefore, the sum of the
solutions is (2 + -1 + -3) = -2

So I think I understand it now.  Thank you for the great help!

Sincerely,
Zahra

```

```
Date: 03/15/2004 at 21:54:26
From: Doctor Ian
Subject: Re: x^3+2x^2-5x-6=0

You're most welcome!  Let us know if you want help with anything else.

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Polynomials
Middle School Factoring Expressions

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