Please Wait... While Loading Full Video#### Class 9 chapter - 2 POLYNOMIALS

### Factorisation of Polynomials

♦ Factorisation of Polynomials

Let us now look at the situation of Example 10 above more closely. It tells us that since the remainder, `q (-1/2) =0 , (2t +1)` is a factor of `q (t)` , i.e, `q(t) = (2t +1) g(t)` for some polynomial `g(t)`. This is a particular case of the following theorem.

`text (Factor Theorem : )` If `p(x)` is a polynomial of degree `n ge 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)`.

`text (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`.

`text (Factor Theorem : )` If `p(x)` is a polynomial of degree `n ge 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)`.

`text (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`.

Q 3230256112

Examine whether `x + 2` is a factor of `x^3 + 3x^2 + 5x + 6` and of `2x + 4`.

Class 9 Chapter 2 Example 11

Class 9 Chapter 2 Example 11

The zero of `x + 2` is `–2`. Let `p(x) = x^3 + 3x^2 + 5x + 6` and `s(x) = 2x + 4`

Then, `p(–2) = (–2)^3 + 3(–2)^2 + 5(–2) + 6`

`= –8 + 12 – 10 + 6`

`= 0`

So, by the Factor Theorem, `x + 2` is a factor of `x^3 + 3x^2 + 5x + 6`.

Again, `s(–2) = 2(–2) + 4 = 0`

So, `x + 2` is a factor of `2x + 4`. In fact, you can check this without applying the Factor

Theorem, since `2x + 4 = 2(x + 2)`.

Q 3250256114

Find the value of k, if `x – 1` is a factor of `4x^3 + 3x^2 – 4x + k`.

Class 9 Chapter 2 Example 12

Class 9 Chapter 2 Example 12

As `x – 1` is a factor of `p(x) = 4x^3 + 3x^2 – 4x + k, p(1) = 0`

Now, `p(1) = 4(1)^3 + 3(1)^2 – 4(1) + k`

So, `4 + 3 – 4 + k = 0`

i.e., `k = –3`

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

`x^2 + lx + m`. You had factorised it by splitting the middle term` lx` as `ax + bx` so that

`ab = m`. Then `x^2 + lx + m = (x + a) (x + b)`. We shall now try to factorise quadratic

polynomials of the type `ax^2 + bx + c`, where `a ≠ 0` and `a, b, c` are constants.

Factorisation of the polynomial `ax^2 + bx + c` by splitting the middle term is as follows:

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

`ax^2 + bx + c = (px + q) (rx + s) = pr x^2 + (ps + qr) x + qs`

Comparing the coefficients of `x^2`, we get `a = pr`.

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 `ax^2 + bx + c`, we have to write b as the sum of two

numbers whose product is `ac`. This will be clear from Example 13.

Q 3280256117

Factorise `6x^2 + 17x + 5` by splitting the middle term, and by using the Factor Theorem.

Class 9 Chapter 2 Example 13

Class 9 Chapter 2 Example 13

(By splitting method) : If we can find two numbers `p` and `q` such that

`p + q = 17` and `pq = 6 × 5 = 30`, then we can get the factors.

So, let us look for the pairs of factors of `30`. 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, `6x^2 + 17x + 5 = 6x^2 + (2 + 15)x + 5`

`= 6x^2 + 2x + 15x + 5`

`= 2x(3x + 1) + 5(3x + 1)`

`= (3x + 1) (2x + 5)`

`text ( Solution 2 : )`(Using the Factor Theorem)

`6x^2 +17 x + 5 = 6 (x^2 +17/6 x + 5/6) = 6 p (x)` say. If `a` and `b` are the zeroes of `p(x)`, then

`6x^2 + 17x + 5 = 6(x – a) (x – b)`. So, `ab = 5/6` . Now , `p (1/2) = 1/4 +17/6 (1/2) +5/6 ≠ 0`. But

`p ( (-1)/3) = 0`. So, ` (x+ 1/3) ` is a factor of `p(x)`. Similarly, by trial, you can find that

`(x+ 5/2)` is a factor of ` p ( x) ` .

Therefore, `6x^2 + 17 x + 5 = 6 (x +1/3) ( x+5/2)`

`= 6 ( (3x+1)/3) ( (2x+5)/2) `

For the example above, the use of the splitting method appears more efficient. However,

let us consider another example.

Q 3200256118

Factorise `y^2 – 5y + 6` by using the Factor Theorem.

Class 9 Chapter 2 Example 14

Class 9 Chapter 2 Example 14

Let `p(y) = y^2 – 5y + 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) = 22 – (5 × 2) + 6 = 0`

So, `y – 2` is a factor of `p(y)`.

Also, `p(3) = 3^2 – (5 × 3) + 6 = 0`

So, `y – 3` is also a factor of `y^2 – 5y + 6`.

Therefore, `y^2 – 5y + 6 = (y – 2)(y – 3)`

Note that `y^2 – 5y + 6` can also be factorised by splitting the middle term `–5y`.

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.

Q 3210356210

Factorise `x^3 – 23x^2 + 142x – 120`.

Class 9 Chapter 2 Example 15

Class 9 Chapter 2 Example 15

Let `p(x) = x^3 – 23x^2 + 142x – 120`

We shall now look for all the factors of `–120`. Some of these are `±1, ±2, ±3,

±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60`.

By trial, we find that `p(1) = 0`. So `x – 1` is a factor of `p(x)`.

Now we see that `x^3 – 23x^2 + 142x – 120 = x^3 – x^2 – 22x^2 + 22x + 120x – 120`

`= x^2(x –1) – 22x(x – 1) + 120(x – 1)` (Why?)

`= (x – 1) (x^2 – 22x + 120)` [Taking `(x – 1)` common]

We could have also got this by dividing `p(x)` by `x – 1`.

Now `x^2 – 22x + 120` can be factorised either by splitting the middle term or by using

the Factor theorem. By splitting the middle term, we have:

`x^2 – 22x + 120 = x^2 – 12x – 10x + 120`

`= x(x – 12) – 10(x – 12)`

`= (x – 12) (x – 10)`

So, `x^3 – 23x^2 – 142x – 120 = (x – 1)(x – 10)(x – 12)`