# Factorisation of Polynomials

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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 n > 1 and a is any real number,then

(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.

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**Example 6**

**Examine whether x + 2 is a factor of **

- The zero of x + 2 is
. Let p(x) = +3x 3 +5x+6x 2 - Let s(x)=2x+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
+3x 3 +5x+6x 2 - 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 4**

- As x – 1 is a factor of p(x) =
- Substitute p(x) = 1
- So p(1) =
- Add the terms, then we get the k=
- Therefore value of k =

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 **ab = m.** Then **constants.**

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Factorisation of the polynomial

Let its factors be (px + q) and (rx + s). Then

Comparing the coefficients of

**Example 8**

**Factorise 6**

- (By splitting method) : If we can find two numbers p and q such that p + q = 17 and pq = 6 x 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.
- We have found the answer.

## 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 = 6. So, to look for the factors of p(y), we look at the factors of 6. - The factors of 6 are 1, 2 and 3.
- Now p(2)
- We get the answer is
- 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