Download The Remainder Theorem

Unit 1.3 The Remainder Theorem
Definition: The Remainder Theorem
If f(x) is a polynomial, then the remainder obtained by dividing f(x) by x-a is f(a).
The Remainder Theorem has a very important role in the evaluating of polynomials at a given
value of x.
The Remainder Theorem begins starts with an unnamed polynomial p(x), where "p(x)" just
means that the variable in the polynomial p is x.
The Theorem then talks about dividing the polynomial by some linear factor x – a, where a is just
some number.
Then, applying long division, you end up with some polynomial answer q(x) (for "quotient
polynomial") and some polynomial remainder r(x).
Example
Let p(x) = x3 – 6x + 7, and let's divide by the linear factor x – 3:
Solution:
x 2  3x  3
x  3 x3  0x2  6x  7
x 3  3x 3
3x 2  6 x  7
3x 2  9 x
3x  7
3x  9
16
So we get x2 + 3x + 3 on top (this is q(x)), with a remainder of 16.
As you recall, from long division of regular numbers from elementary school, that your remainder (if
there is one) has to be less than whatever you divided by. In polynomial terms, since we're dividing
by a linear factor (that is, a factor in which the degree on x is just “1”), then the remainder must be
just a constant. That is, when you divide by "x – a", your remainder will just be some number.
The Remainder Theorem then reminds you about the connection between division and multiplication.
For instance, since 15 ÷ 3 = 5, then 5 × 3 = 15. If you get a remainder, you do the multiplication
and then add the remainder back in. For instance, since 14 ÷ 5 = 2 with R 4, then 14 = 5 × 2 + 4.
This process works the same way with polynomials. That is:
p  x
 q  x  with the remainder r(x)
xa
This can be written as
p(x) = (x – a)q(x) + r(x).
Our example verifies this property:
Since
x3  6 x  7
 x 2  3x  3 with a remainder of 16
x3
Then x3 – 6x + 7 = (x-3)( x2 + 3x + 3)+16
Then the Remainder Theorem says that we can rewrire the polynomial in terms of the divisor, and
then evaluate the polynomial at x = a. But when x = a, the factor "x – a" is just zero! Then:
p(a) =
=
=
=
(a – a)q(a) + r(a)
(0)q(a) + r(a)
0 + r(a)
r(a)
But remember that the remainder term r(a) is just a number! So the value of the polynomial p(x)
at x = a is the same as the remainder when you divide p(x) by x – a. In terms of the above
example:
p(3) = (3 – 3)((3)2 + 3(3) + 3) + 16
= (0)(9 + 9 + 9) + 16
= 0 + 16
= 16
Remember that, when you are dividing by a linear factor, you can instead use synthetic division,
which is much quicker.
In our example, we would get:
3
1 0 6 7
3
9 9
1 3
3 16
Did you notice that the last entry in the bottom row is 16, which is the same as the remainder from
the long division (as expected) and also the value of x 3 – 6x + 7 at x = 3.
Example
What is the remainder when
x3  4 x 2  5x  1 is divided by x  1
Solution:
Let
f  x   x 3  4 x 2  5x  1
The remainder when
f  x  is divided by x  1 is f  1
f  1   1  4  1  5  1  1
3
2
 1  4  5  1
 11
The remainder is -11.
Example
When
x3  3x 2  kx  10 is divided by x  5 , the remainder is 30. find the value of k.
Solution:
Let
f  x   x3  3x 2  kx  10
The remainder when
f  x  is divided by x  5 is f  5 .
f  5   5  3  5  5k  10
3
2
 125  75  5k  10
 210  5k
Now since the remainder is 30,
210  5k  30
5k  180
k  36
The advantage of the remainder theorem is that we can determine if a value is a factor by
checking if the remainder is zero.
The Factor Theorem
The Factor Theorem is based upon the properties of the Remainder Theorem. If f(a)=0 then the
remainder is 0 and x-a is a factor. It can also be noted that you can go the other way, if x-a is a
factor, then f(a)=0.
Example
Fully factor
2 x3  7 x 2  2 x  3
Solution:
Examine the integer factors of 3
f  1  2  1  7  1  2  1  3
3
2
 2  7  2  3
0
Therefore x+1 is a factor of
2 x3  7 x 2  2 x  3
Now by synthetic division
1
2 7 2 3
2 5
3
2 5 3
0
Which is
2 x 2  5x  3
Now factoring
Thus
2 x 2  5x  3 we have  2 x  1 x  3
2 x3  7 x 2  2 x  3   x  1 2 x  1 x  3 .