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:
x2 3 x 3 x3 x3 0 x 2 6 x 7 x33 x 3 3x2 6 x 7 3x2 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,then13=5×2+ 4. This process works the same way with polynomials. That is: p x q xwith the remainder r(x) x a
This can be written as p(x) = (x – a)q(x) + r(x).
Our example verifies this property:
x3 6 x 7 Since x2 3 x 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:
1 0 6 7 3 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 x3 – 6x + 7 at x = 3.
Example
What is the remainder when x34 x 2 5 x 1 is divided by x 1
Solution: Let f x x34 x 2 5 x 1 The remainder when f xis divided by x 1 is f 1
f 1 13 4 1 2 5 1 1 1 4 5 1 11
The remainder is -11.
Example
When x33 x 2 kx 10 is divided by x 5, the remainder is 30. find the value of k.
Solution: 3 2 Let f x x3 x kx 10
The remainder when f xis divided by x 5 is f 5.
f5 53 3 5 2 5 k 10 125 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 2x3 7 x 2 2 x 3
Solution:
Examine the integer factors of 3 f 1 2 13 7 1 2 2 1 3 2 7 2 3 0
Therefore x+1 is a factor of 2x3 7 x 2 2 x 3
Now by synthetic division
2 7 2 3 1 2 5 3 2 5 3 0
Which is 2x2 5 x 3
Now factoring 2x2 5 x 3 we have 2x 1 x 3
Thus 2x3 7 x 2 2 x 3 x 1 2 x 1 x 3.