NCERT Class 9 Mathematics Textbook for Blind and Visually Impaired Students made Screen Readable by Professor T K bansal.

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.

The Factor Theorem :

If p(x) is a polynomial of degree n greater than or equal 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 some polynomial g(x).
In this case, p(a) = (a − a) g(a) = 0.

Example 11

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

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, we can check this without applying the Factor Theorem, since 2x + 4 = 2(x + 2).

Example 12

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

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. We are already familiar with the factorisation of a quadratic polynomial like x^2 + lx + m. We 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 x2 + (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.

Example 13

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

Solution 1 : (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)

Solution 2 :

(Using the Factor Theorem)
6x^2 + 17x + 5 = 6 (x^2 +17/6 +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. Let us look at some possibilities for a and b. They could be ±½, ±1/3, ±5/3, ±5/2. ±1.
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 + 17x + 5 = 6 (x+1/3)(x+ 5/2)
=6 [(3x+1)/3] [(2x+5)/2]
= (3x + 1) (2x + 5)

For the example above, the use of the splitting method appears more efficient. However, let us consider another example.

Example 14

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

Solution :

Let p(y) = y^2 − 5y + 6. Now, if p(y) = (y − a) × (y − b), we know that the constant term will be a b. So, a b = 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) = 2^2 − (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 we will see in the following example.

Example 15

Factorise x^3 − 23x^2 + 142x − 120.

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)

EXERCISE 2.4

Q1. Determine which of the following polynomials has (x + 1) a factor :
(i) x^3 + x^2 + x + 1
(ii) x^4 + x^3 + x^2 + x + 1
(iii) x^4 + 3x^3 + 3x^2 + x + 1
(iv) x^3 − x^2 − (2 + √2)x + √2

A1.
(x + 1) is a factor of (i), but not the factor of (ii), (iii) and (iv).

Q2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x3 + x2 − 2x − 1, g(x) = x + 1
(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
(iii) p(x) = x3 − 4x2 + x + 6, g(x) = x − 3

A2.
(i) Yes
(ii) No
(iii) Yes

Q3. Find the value of k, if x − 1 is a factor of p(x) in each of the following cases:
(i) p(x) = x^2 + x + k
(ii) p(x) = 2x^2 + kx + √2
(iii) p(x) = kx^2 − 2 x + 1
(iv) p(x) = kx^2 − 3x + k

A3.
(i) −2
(ii) −(2 + √2)
(iii) √2 − 1
(iv)3/2

Q4. Factorise :
(i) 12x^2 − 7x + 1
(ii) 2x^2 + 7x + 3
(iii) 6x^2 + 5x − 6
(iv) 3x^2 − x − 4

A4.
(i) (3x – 1) (4x – 1)
(ii) (x + 3) (2x + 1)
(iii) (2x + 3) (3x – 2)
(iv) (x + 1) (3x – 4)

Q5. Factorise :
(i) x^3 − 2x^2 − x + 2
(ii) x^3 − 3x^2 − 9x − 5
(iii) x^3 + 13x^2 + 32x + 20
(iv) 2y^3 + y^2 − 2y − 1

A5. (i) (x – 2) (x – 1) (x + 1)
(ii) (x + 1) (x + 1) (x – 5)
(iii) (x + 1) (x + 2) (x + 10)
(iv) (y – 1) (y + 1) (2y + 1)