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Factorisation of Polynomials and Factor Theorem - Polynomials, Class 9, Mathematics PDF Download

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), then p(a) = 0 and

(ii) p(a) = 0 then x – a is a factor of p(x).

REMARK

  1. (x + a) is a factor of p(x) ⇔ p(–a) = 0
  2.  (ax – b) is a factor of p(x) ⇔ NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 = 0
  3.  (ax + b) is a factor of p(x) ⇔NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9= 0
  4. (x – a) (x – b) is a factor of a polynomial p(x) ⇔ p(a) = 0, p(b) = 0

Ex. Use the factor theorem to determine whether x – 1 is a factor of

(a) x3 + 8x2 – 7x – 2

(b) 2√2 x3 + 5√2 x2 – 7√2

Sol. (a) Let p(x) = x3 + 8x2 – 7x – 2

By using factor theorem, (x – 1) is a factor of p(x) only when p(1) = 0

p(1) = (1)3 + 8(1)2 – 7(1) – 2 = 1 + 8 – 7 – 2 = 9 – 9 = 0

Hence (x – 1) is a factor of p(x).

(b) Let p(x) = 2√2x3 + 5√2x2 – 7√2

By using factor theorem, (x – 1) is a factor of p(x) only when p(1) = 0

p(1) = 2√2 (1)3 + 5√2 (1)2 – 7√2 = 2√2 + 5√2 – 7√2 = 7√2 – 7√2 = 0

Hence (x – 1) is a factor of p(x).

 

Ex. Without actual division, prove that the polynomial 2x3 + 13x2 + x – 70 is exactly divisible by x – 2.

Sol. The polynomial p(x) = 2x3 + 13x2 + x – 70 is exactly divisible by x – 2 means that x – 2 is a factor of

p(x) = 2x3 + 13x2 + x – 70.

Now p(2) =2(2)3 + 13(2)2 + 2 – 70 = 16 + 52 + 2 – 70 = 0

∴ By factor theorem, x – 2 is a factor of p(x)

i.e. p(x) = 2x3 + 13x2 + x – 70 is exactly divisible by x – 2.

FACTORISING OF POLYNOMIAL OF HIGHER DEGREE

  1. Obtain the polynomial p(x)
  2. Obtain the constant term in p(x) and find its all possible factors. For example, in the polynomial x4 + x3 – 7x2 – x + 6 the constant term is 6 and its factors are ±1, ±2, ±3, ±6.
  3. Take one of the factors, say a and replace x by it in the given polynomial. If the polynomial reduces to zero, then (x – a) is a factor of polynomial.
  4. Obtain the factors equal in number to the degree of polynomial. Let these are (x – a), (x – b), (x – c). ....
  5. Write p(x) = k (x – a) (x – b) (x – c) ..... where k is constant.
  6. Substitute any value of x other than a, b, c .......... and find the value of k.

Ex. Using factor theorem, factorize the polynomial x3 – 6x2 + 11x – 6.
Sol. Let f(x) = x3 – 6x2 + 11x – 6

The constant term in f(x) is equal to –6 and factors of – 6 are ±1, ±2, ±3, ±6. Putting x = 1 in f(x), we have

f(1) = (1)3 – 6 × (1)2 + 11 × 1 – 6 = 1 – 6 + 11 – 6 = 0     ∴        (x – 1) is a factor of f(x)

Similarly, x – 2 and x – 3 are factors of f(x). Since f(x) is a polynomial of degree 3. So, it can not have more than three linear factors.

Let f(x) = k (x – 1) (x – 2) (x – 3)

Then,x3 – 6x2 + 11x – 6 = k(x – 1) (x – 2) (x – 3)

Putting x = 0 on both sides, we get – 6 = k (0 – 1) (0 – 2) (0 – 3) ⇒   –6 = – 6 k      ⇒   k = 1

Putting k = 1 in f(x) = k(x – 1) (x – 2) (x – 3), we get f(x) = (x – 1) (x – 2) (x – 3)

Hence, x3 – 6x2 + 11x – 6 = (x – 1) (x – 2) (x – 3)

Ex. Using factor theorem, factorize the polynomial x4 + x3 – 7x2 – x + 6.

Sol. Let f(x) = x4 + x3 – 7x2 – x + 6 the factors of constant term 6 are ±1, ±2, ±3 and ±6

Now, f(1) = 1 + 1 – 7 – 1 + 6 = 8 – 8 = 0

⇒ (x – 1) is a factor of f(x)

f(–1) = 1 – 1 – 7 + 1 + 6 = 8 – 8 = 0     ⇒     (x + 1) is a facor of f(x)

f(2) = 24 + 23 – 7 × 22 – 2 + 6 = 16 + 8 – 28 – 2 + 6 = 0

⇒ x – 2 is a factor of f(x)

f(–2) = (–2)4 + (–2)3 – 7(–2)2 – (–2) + 6 = 16 – 8 – 28 + 2 + 6 = – 12    ≠ 0

⇒ x + 2 is not a factor of f(x)

f(–3) = (–3)4 + (–3)3 – 7(–3)2 – (–3) + 6 = 81 – 27 – 63 + 3 + 6 = 90 – 90 = 0

⇒ x + 3 is a factor of f(x).

Since f(x) is a polynomial of degree 4. So, it cannot have more than 4 linear factors

Thus, the factor of f(x) are (x – 1), (x + 1), (x – 2) and (x + 3).

Let f(x) = k(x – 1) (x + 1) (x – 2) (x + 3)

⇒ x4 + x3 – 7x2 – x + 6 = k (x – 1) (x + 1) (x – 2) (x + 3) ... (i)

Putting k = 0 on both sides, we get

6 = k (–1) (1) (–2) (3)        ⇒     6 = 6 k          ⇒          k = 1

Substituting k = 1 in (i), we get

x4 + x3 – 7x2 – x + 6 = (x – 1) (x + 1) (x – 2) (x + 3)

 

COMPETITION WINDOW

ZERO OF A QUADRATIC POLYNOMIAL
Symmetric function : An algebraic expression in α and β, which remains unchanged, when α and β are interchanged is known as symmetric function in α and β.
For example, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9etc. are symmetric functions. Symmetric function is to be expressed in terms of (α+ β) and αβ. So, this can be evaluated for a given quadratic equation.

Some useful relations involving α and β :

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Factorisation of Polynomials and Factor Theorem - Polynomials, Class 9, Mathematics

Factorisation of Polynomials and Factor Theorem - Polynomials, Class 9, Mathematics

Factorisation of Polynomials and Factor Theorem - Polynomials, Class 9, Mathematics

Ex. If α and β are the zeroes of the polynomial ax2 + bx + c. Find the value of
(i) α – β

(ii) α2 + β2
Sol.
Since α and β are the zeroes of the polynomial ax2 + bx + c.

Factorisation of Polynomials,Class IX,Important Notes,Maths,Polynomials,Factor Theorem

(i) (α – β)2 = (α + β)2 – 4αβ

Factorisation of Polynomials,Class IX,Important Notes,Maths,Polynomials,Factor Theorem

Factorisation of Polynomials,Class IX,Important Notes,Maths,Polynomials,Factor Theorem

(ii) α2 + β2 = α2 + β2 + 2αβ – 2αβ = (α + β)2 – 2αβ

Factorisation of Polynomials,Class IX,Important Notes,Maths,Polynomials,Factor Theorem

Ex. If α and β are the zeroes of the quadratic polynomial ax2 + bx + c. Find the value of

(i) α2β2

(ii) α3 + β3.
Sol.
Since α and β are the zeroes of ax2 + bx + c

Factorisation of Polynomials,Class IX,Important Notes,Maths,Polynomials,Factor Theorem

(i) α2β2 = (α + β) (α – β)

Factorisation of Polynomials,Class IX,Important Notes,Maths,Polynomials,Factor Theorem

Factorisation of Polynomials,Class IX,Important Notes,Maths,Polynomials,Factor Theorem

(ii) α3 + β3 = (α + β) (α2 + β2 – αβ) = (α + β) [(α2 + β2 + 2αβ) – 3αβ] = (α + β) [(α + β)2 – 3αβ] =

Factorisation of Polynomials and Factor Theorem - Polynomials, Class 9, Mathematics

Factorisation of Polynomials and Factor Theorem - Polynomials, Class 9, Mathematics

 

TO FORM A QUADRATC POLYNOMIAL WITH THE GIVEN ZEROES

Let the zeroes of a quadratic polynomial be α and β.
∴ x = α,               x = β
   x – α = 0,          x – β = 0
The obviously the quadratic polynomial is
(x – α) (x – β)
i.e., x2 – (α + β) x + αβ
x2 – (Sum of the zeroes) x + Product of the zeroes

Ex. Form the quadratic polynomial whose zeroes are 4 and 6.
Sol. Sum of the zeroes = 4 + 6 = 10
Product of the zeroes = 4 × 6 = 24
Hence the polynomial formed = x2 – (sum of zeroes) x + Product of zeroes = x2 – 10x + 24

Ex. Form the quadratic polynomial whose zeroes are –3, 5
Sol. Here, zeroes are – 3 and 5.
Sum of the zeroes = –3 + 5 = 2
Product of the zeroes = (–3) × 5 = –15
Hence the polynomial formed = x2 – (sum of zeroes) x + Product of zeroes = x2 – 2x – 15

The document Factorisation of Polynomials and Factor Theorem - Polynomials, Class 9, Mathematics is a part of Class 9 category.
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FAQs on Factorisation of Polynomials and Factor Theorem - Polynomials, Class 9, Mathematics

1. What is a polynomial?
Ans. A polynomial is a mathematical expression consisting of variables, coefficients, and exponents, which are combined using the basic mathematical operations of addition, subtraction, multiplication, and division. Examples of polynomials include x^2 + 2x + 3, 2x^3 - 4x^2 + x - 5, etc.
2. What is factorisation of polynomials?
Ans. Factorisation of polynomials is the process of expressing a polynomial as a product of two or more simpler polynomials. It is similar to finding the prime factors of a number. By factorising a polynomial, we can easily solve equations, find the roots of the polynomial, and simplify complex expressions.
3. What is the Factor Theorem?
Ans. The Factor Theorem is a theorem in algebra that states that if a polynomial f(x) is divided by x-a (where a is a constant), and the remainder is zero, then x-a is a factor of f(x). In other words, if f(a) = 0, then (x-a) is a factor of f(x). The Factor Theorem is useful in finding the factors of a polynomial, and in finding the roots of a polynomial equation.
4. How do you factorise a quadratic polynomial?
Ans. To factorise a quadratic polynomial of the form ax^2 + bx + c, we need to find two numbers that multiply to give ac and add to give b. These numbers are called the factors of ac. Once we have found the factors of ac, we can rewrite the quadratic polynomial as (px + q)(rx + s), where p and r are the factors of a, q and s are the factors of c, and ps+qr=b.
5. What are the practical applications of factorisation of polynomials?
Ans. Factorisation of polynomials has a wide range of practical applications in fields such as engineering, physics, economics, and computer science. It is used to solve polynomial equations, find the roots of a polynomial, simplify complex expressions, and solve problems related to optimization, control systems, cryptography, and data compression. Factorisation of polynomials is also used in the study of number theory, algebraic geometry, and algebraic topology.
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