THE FACTOR AND REMAINDER THEOREMS
1. Algebraic Division
First, a reminder of long division with numbers.
6 / 511 / 7 / 2 / 1
6 / 6
6 / 1
5 / 5
6
So = 65 remainder 6, or 721 = (65 × 11) + 6.
We say 65 is the quotient, 11 is the divisor, and 6 is the remainder.
Example 1 : Divide 50432 by 23 using long division.
2 / 1 / 9 / 223 / 5 / 0 / 4 / 3 / 2
4 / 6
4 / 4
2 / 3
2 / 1 / 3
2 / 0 / 7
6 / 2
4 / 6
1 / 6
So = 2192 remainder 16, or 50432 = (2192 × 23) + 16.
We now apply the same principles to algebra.
Example 2 : Divide by.
5x / − / 1x / + / 1 / 5x2 / + / 4x / − / 1
5x2 / + / 5x
− / x / − / 1
− / x / − / 1
0
So . This can be easily checked by expanding .
Example 3 : Simplify .
x3 / − / 2x2 / − / 8x / + / 1x / − / 3 / x4 / − / 5x3 / − / 2x2 / + / 25x / − / 3
x4 / − / 3x3
− / 2x3 / − / 2x2 / + / 25x / − / 3
− / 2x3 / + / 6x2
− / 8x2 / + / 25x / − / 3
− / 8x2 / + / 24x
x / − / 3
x / − / 3
0
So
Example 4 : Find the missing factor in this identity :
2x3 / + / 3x2 / − / x / + / 62x / − / 3 / 4x4 / − / 0x3 / − / 11x2 / + / 15x / − / 18
4x4 / − / 6x3
6x3 / − / 11x2 / + / 15x / − / 18
6x3 / − / 9x2
− / 2x2 / + / 15x / − / 18
− / 2x2 / + / 3x
12x / − / 18
12x / − / 18
0
Notice the importance of including in the division.
Therefore we have
Example 5 : Divide by.
3x3 / − / 2x2 / + / 6x / − / 6−x / + / 2 / −3x4 / + / 8x3 / − / 10x2 / + / 18x / − / 14
−3x4 / + / 6x3
2x3 / − / 10x2 / + / 18x / − / 14
2x3 / − / 4x2
− / 6x2 / + / 18x / − / 14
− / 6x2 / + / 12x
6x / − / 14
6x / − / 12
−2
We have a remainder of −2. This means that
Example 6 : Divide by.
x2 / − / 3x / + / 15x2 / + / 3x / − / 1 / x4 / + / 0x3 / + / 5x2 / + / 0x / − / 2
x4 / + / 3x3 / − / x2
−3x3 / + / 6x2 / + / 0x / − / 2
−3x3 / − / 9x2 / + / 3x
15x2 / − / 3x / − / 2
15x2 / + / 45x / − / 15
−48x / + / 13
This time the remainder is. This means that
C2 p8 Ex 1B, p9 Ex 1C
2. The Factor Theorem
Consider the prime factorisation of 24.
All the factors of 24 (except 1) can be made up from these prime factors. For example , and. In the same way the factors of the function
are , , and . These are known as linear factors, but there are also quadratic factors such as and .
From the definition of , we can see that
because any value of x which makes any linear factor equal to zero will automatically make equal to zero.
We can generalise this into the factor theorem, which states that...
Note the connection with what we learned in unit C1 about where the graph of a function crosses the x-axis.
We can prove the factor theorem formally by supposing that is a factor of , in which case we can write
where is some other function of x. Now let ...
...and we have the factor theorem.
The factor theorem can help us begin to factorise cubic expressions.
Example1:Factorise completely the expression . Hence solve the equation .
We see that the factorised expression will be of the form , and since , the possible values of a, b and c are ±1, ±2, ±4 and ±8. We work our way through these options, using the factor theorem. Start with the simplest numbers first!
so is not a factor.
so is a factor.
We could now divide byusing the methods in the previous section.
However, it is simpler to write
Theand 8 in the quadratic factor are obvious. To find k, the coefficient of x, we can either compare coefficients of on both sides of the identity...
...or compare coefficients of x...
Either way we have...
...the quadratic being easy to factorise.
To solve our equation,
There is a tabular version of this method which does not involve using k.
/ 8 / / 8 / / / 81 / / 8 / 1 / / / 8 / 1 / / / 8
The first table shows what we already know. We then fill in the number of and x needed in the middle column, and finally work out what the column heading must be.
Example 2 : Factorise completely the expression .
so is not a factor.
so is not a factor.
so is not a factor.
so is a factor.
So we can write
Comparing coefficients of
And so
Again, we could have used the tabular method.
/ –6 / / –6 / / / –62 / / –12 / 2 / / / –12 / 2 / / / –12
Example3:Prove thatis a factor of . Hence solve the equation .
By the Factor Theorem, if is a factor then . Note : a lot of students do an unnecessary long division to prove is a factor of .
Therefore is a factor of . Now
Comparing coefficients of
And so
Solving the equation
At this point, consider the function
We can see that
The last result is because x must be to make the factor equal to zero. More generally x must be to make the factorequal to zero. A complete version of the factor theorem is therefore
Example 4 : Factorise .
Possible linear factors are, , and . so is not a factor.
so is not a factor.
so is not a factor.
so is not a factor.
so is not a factor.
so is not a factor.
so is not a factor.
so is a factor.
Comparing coefficients of
And so
The quadratic does not factorise, but then we know that already, as we exhausted our list of linear factors.
Example5:The polynomial is exactly divisible by and . Find the values of p and q and hence factorise .
Note : in examinations, a lot of candidates latch on to the word ‘divisible’ and attempt an algebraic division. This has messy and time-consuming consequences, since the values of p and q are not known. The correct method is to use the factor theorem.
Since is a factor,
Since is a factor,
Subtracting the first simultaneous equation from the second,
We now have
The last factor can be written down immediately, as it is the only one that will give and 8 in the expansion.
Example6:. Given that and are factors of , find the values of a and b. Hence factorise completely.
Subtracting,
And so
Comparing coefficients of, we have
And so.
C2 p12 Ex 1D
3. The Remainder Theorem
For example, if we divide by the remainder, according to the theorem, should be .
4x2 / + / 10x / + / 13x / – / 2 / 4x3 / + / 2x2 / – / 7x / + / 6
4x3 / – / 8x2
10x2 / – / 7x / + / 6
10x2 / – / 20x
13x / + / 6
13x / – / 26
32
So the theorem works (in this case at least!)
Similarly, if we divide by , the remainder should be .
3x3 / – / 2x2 / + / 12x / – / 36x / + / 3 / 3x4 / + / 7x3 / + / 6x2 / + / 0x / – / 10
3x4 / + / 9x3
– / 2x3 / + / 6x2 / + / 0x / – / 10
– / 2x3 / – / 6x2
12x2 / + / 0x / – / 18
12x2 / + / 36x
– / 36x / – / 18
– / 36x / – / 108
98
Once again the theorem works.
To see why this is, consider the first division. We can write the answer like this...
This identity is always true, and so must be true for any value of x we choose. Let’s pick. Substituting this in gives us, which is the remainder!
For the second division, we have
Choosing gives us , which is again the remainder.
We prove the remainder theorem as follows.
When a polynomial is divided by, suppose the quotient is and the remainder r. Then we have
Substituting into this equation,
Notice that the factor theorem is a special case of the remainder theorem. If is a factor of , then and hence.
The more general case of the remainder theorem is
We prove this in a similar way. Suppose
Substituting into this equation,
Example 1 : Find the remainder when is divided by .
A simple rule to find the value of x to substitute into is to use the value of x which makes the factor equal to zero. In this case makes equal to zero.
Checking by long division,
2x2 / – / 3x / + / 23x / + / 2 / 6x3 / – / 5x2 / + / 0x / + / 13
6x3 / + / 4x2
– / 9x2 / + / 0x / + / 13
– / 9x2 / – / 6x
6x / + / 13
6x / + / 4
9
Example2:The remainder obtained when is divided by is equal to the remainder obtained when the same expression is divided by. Find the value of a.
We have
Example3:The polynomial is exactly divisible by and has remainder 45 when divided by . Find the values of the constants A and B. Hence solve the equation .
Now
And also
Adding equations,
And so
Comparing coefficients of,
Now
The solutions of are therefore .
C2 p14 Ex 1E Topic Review : Factor and Remainder Theorems