Factorisation of Polynomials
Let us now look at the situation of Example 10 above more closely. It tells us that since the remainder, q
Factor Theorem :
If p(x) is a polynomial of degree
(i) x – a is a factor of p(x), if p(a) = 0, and
(ii) p(a) = 0, if x – a is a factor of p(x).
Proof:
By the Remainder Theorem, p(x)=(x – a) q(x) + p(a).
(i) If p(a) = 0, then p(x) = (x – a) q(x), which shows that x – a is a factor of p(x).
(ii) Since x – a is a factor of p(x), p(x) = (x – a) g(x) for same polynomial g(x). In this case, p(a) = (a – a) g(a) = 0.
Example 6
Examine whether x + 2 is a factor of
- The zero of x + 2 is
. Let p(x) = x 3 + 3 x 2 + 5 x + 6 - Let s(x) =
2 x + 4 - Then substitute
x = − 2 value in the factor of p(x) - Calculate the factor and add it.
- We get the answer as
- So, by the Factor Theorem, x + 2 is a factor of
x 3 + 3 x 2 + 5 x + 6 - Again, substitute the
x = − 2 value in s(x) - We get the answer as
- So, x + 2 is a factor of 2x + 4. In fact, you can check this without applying the Factor Theorem, since 2x + 4 =
(x + 2).
Example 7
Find the value of k, if x – 1 is a factor of
- As x – 1 is a factor of p(x) we get:
- Substitute p(x) = 1
- So p(1) becomes
- Add the terms, then we get the k=
- Therefore value of k has been obtained.
We will now use the Factor Theorem to factorise some polynomials of degree 2 and 3.
You are already familiar with the factorisation of a quadratic polynomial like
We shall now try to factorise quadratic polynomials of the type
Factorisation of the polynomial
Let its factors be (px + q) and (rx + s). Then
Comparing the coefficients of
Similarly, comparing the coefficients of x, we get b = ps + qr.
And, on comparing the constant terms, we get c = qs.
This shows us that b is the sum of two numbers ps and qr, whose product is (ps)(qr) = (pr)(qs) = ac.
Therefore, to factorise
Example 8
Factorise
- (By splitting method): If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 =
, then we can get the factors. - Some are 1 and 30, 2 and 15, 3 and 10, 5 and 6. Of these pairs, 2 and 15 will give us p + q = 17.
- So
6 x 2 + 17 x + 5 we can apply the 15,2 within the 17x - Multiply the values with x
- Split the terms with factors.
- Separate the terms: (
+ 1) + ( + 5) - We have found the answer.
Solution 2: (Using the Factor Theorem)
Let's say:
Assume: If a and b are the zeroes of p(x), then:
6p(x) = 6(x-a)(x-b).
So, ab =
Let us look at some possibilities for a and b.
They could be
Now,
But
So,
Similarly, by trial, you can find that
Therefore,
=
Example 9
Factorise
- Let p(y) =
y 2 − 5 y + 6 Now, if p(y) = (y – a) (y – b), you know that the constant term will be ab. So, ab =. So, to look for the factors of p(y), we look at the factors of . - The factors of 6 are 1,
and . - Now p(2) becomes
- We get the answer is equal to
- So, y – 2 is a factor of p(y).
- Also, p(3) =
- So, y – 3 is also a factor of
y 2 − 5 y + 6 - Therefore,
y 2 − 5 y + 6 = (y – 2)(y – 3)
Note that
Now, let us consider factorising cubic polynomials.
Here, the splitting method will not be appropriate to start with. We need to find at least one factor first, as you will see in the following example.
Example 10 :