Factorization of Polynomials Notes | Study Mathematics (Maths) Class 9 - Class 9

Factor theorem:

Let p(x) be a polynomial of degree n ≥ 1 and a be any real number such that,
(i) p(a) = 0, then (x − a) is a factor of p(x) conversely,
(ii)If (x − a) is a factor of p(x), then p(a) = 0

Factor Theorem 

Proof:
(i) Given p(a) = 0
Now, suppose p(x) is divided by (x − a), then quotient is g(x).
By remainder theorem, when p(x) is divided by (x − a), then remainder is p(a).
Dividend = Divisor × Quotient + Remainder
∴ p(x) = (x − a). g(x) + p(a)
p(x) = (x − a). g(x) + 0 [∵ p(a) = 0]
So, (x − a) is a factor of p(x)
(ii) Let (x − a) be a factor of p(x)
On dividing p(x) by (x − a), let g(x) be the quotient.
∴ p(x) = (x − a). g(x)
On putting x = a, we get
p(a) = (a − a). g(a)
p(a) = 0. g(a)
p(a) = 0
Thus, if (x − a) is a factor of p(x), then p(a) = 0

Examine whether x + 1 is a factor of p (x) = x³ + 3x3 + 5x + 6
Let p(x) = x³ + 3x² + 5x + 6
By factor theorem, (x − a) is a factor of p(x) if p(a) = 0.

Therefore, in order to check that (x + 1) is a factor of p(x), it is sufficient to check that if p(−1) = 0 then it is a factor otherwise not.

p(x) = x³ + 3x² + 5x + 6
p(– 1) = (– 1)³+ 3(– 1)² + 5(– 1) + 6
p(– 1) = −1 + 3 − 5 + 6
p(– 1) = −6 + 9
p(– 1) = 3 ≠ 0
Thus, x + 1 is NOT a factor of p(x) = x3 + 3x2 + 5x + 6

Find the value of a if x-a is a factor of x³ - a2x + x + 2
Let p(x) = x³ − a²x + x + 2
By factor theorem, (x − a) is a factor of p(x) if p(a) = 0
Now, p(a) = a³ − a² × a + a + 2
a³ − a² × a + a + 2 = 0
a³ − a³ + a + 2 = 0
a + 2 = 0
a = 0 − 2
a = −2
Hence, (x − a) is a factor of the given polynomial, if a = −2

Factorisation of quadratic polynomial

Quadratic polynomial is represented as ax² + bx + c, where a, b, c are constants and a ≠ 0. It can be factorised by different methods.

• By splitting the middle term
• By using factor theorem

(i) By splitting middle term

Let the factor of quadratic polynomial ax² + bx + c be (px + q) and (rx + s). Then

ax² + bx + c = (px + q)(rx + s)

= prx² + (ps + qr)x + qs
Now,
Comparing x², x and constant terms, we get
a = pr, b = ps + qr and c = qs
Here, b is the sum of two numbers ps and qr, whose product is (ps)(qr) = (pr)(qs) = ac
Thus, to factories ax² + bx + c , write b as the sum of two numbers, whose product is ac.
Factorizing 2x² + 7x + 3 by splitting the middle term
Given polynomial is 2x² + 7x + 3 .
On comparing with ax² + bx + c, we get
a = 2, b = 7 and c = 3
Now ac = 2 × 3 = 6
So, all possible pairs of factors 6 is 1 and 6, 2 and 3.
Pair 1 and 6 give 1 + 6 = 7 = b

∴ 2x² + 7x + 3 = 2x² + (1 + 6)x + 3
= 2x² + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)

(ii) By using factor theorem
Write the given polynomial p(x) = ax? + bx + c in the form

i.e. firstly make the coefficient of x2 equal to 1 if it is not 1.

Factories x² - 5x +  6 by using factor theorem.
Let given polynomial be f(x) = x2 − 5x + 6
Here, the coefficient of x² is 1, so we do not need to write it in the form of a. g(x).
Now, constant term is 6 and all factors of 6 are ±1, ±2, ±3 and ± 6
At x = 2,
f(2) = (2)2 − 5(2) + 6
f(2) = 4 − 10 + 6
f(2) = −6 + 6
f(2) = 0
At x = 3,
f(3) = (3)² − 5(3) + 6
f(3) = 9 − 15 + 6
f(3) = −6 + 6
f(3) = 0

Hence, (x − 2) and (x − 3) are the factors of the given quadratic polynomial.

Factorisation of a cubic polynomial

We use the following steps to factorise a cubic polynomial,
Step I: Write the given cubic polynomial, p(x) = ax³+ bx² + cx + d as,

i.e., first make the coefficient of x= equal to 1 if it is not 1, then find the constant term.

Step II: Find all the possible factors of constant term (d/a) of g(X).

Step III: Check at which factor of constant term, p(x) is zero and get one factor of p(x), (i.e., x − α)
Step IV: Write p(x) as the product of this factor (x − α) and a quadratic polynomial i.e., p(x) = (x − α) (a₁x² + b₁x + c₁)
Step V: Apply splitting method of middle term or factor theorem in quadratic polynomial to get the other two factors. Thus, we get all the three factors of given the cubic polynomial.
Using factor theorem, factorise x³ + 13x² + 32x + 20
Let p(x) = x³ + 13x² + 32x + 20
Here the constant term = 20 and the coefficient of x3 is 1.
All possible factors of 20 are ±1, ±2, ±4, ±5, ±10, and ± 20
At x = −1,
p(x) = x³ + 13x² + 32x + 20
p(−1) = (−1)³ + 13(−1)² + 32(−1) + 20
= −1 + 13 − 32 + 20
= 33 − 33
= 0

So, we find p(−1) = 0 So, (x + 1) is a factor of p(x)

We can write the given polynomial x³+ 13x² + 32x + 20 as, x³ + x² + 12x² + 12x + 20x + 20
x²+x²+12x²+12x+20x+20
= x²(x+1)+12x(x+1)+20(x+1)
= (x+1)(x²+12x+20) ........ (i) [Taking (x + 1) common]

Now, x² + 12x + 20 can be factorised by splitting the middle term,
we get,
x² + 12x + 20
= x² + 10x + 2x + 20
= x(x + 10) + 2(x + 10)
= (x + 2)(x + 10) ...........(ii)
From equation (i) and (ii) we get,
x³+ 13x²+ 32x + 20 = (x + 1)(x + 2)(x + 10)
If p(y) = y³ −4y² + y + 6 then show that p(3) = 0 and hence factorise p(y).
Given: p(y) = y³ − 4y² + y + 6...... (i)
Put y = 3 in equation (i), we get
p(3) = (3)³ − 4(3)² + 3 + 6
= 27 − 36 + 3 + 6 = 0
Since, p(3) = 0, therefore y – 3 is a factor of p(y).
∴ p(y) = (y − 3)(y² − y − 2)
∴ p(y) = (y − 3)(y² − 2y + y − 2)
[∵ −2 + 1 = −1 and − 2 × 1 = −2]
= (y − 3)[y(y − 2) + 1(y − 2])
= (y − 3)(y + 1)(y − 2)