Class 9 Factorisation of Polynomials

### Topic Covered

♦ 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.

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
Solution:

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
Solution:

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
Solution:

(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
Solution:

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
Solution:

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)