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POLYNOMIALS
Polynomials consists of Constants (fixed numeric values) and variables (assigned different numerical values). It consists of combination of two or more terms connected by +, -, is called an algebraic expression and each part is called Term in which variables involved having only non-negative integral exponents is called Polynomial denoted by p(x), q(x), r(x) and s(x) etc.
CLASSIFICATION OF POLYNOMIALS
Polynomials are classified on the basis of Degree and Number of Terms
On the basis of Degree: Highest power of the variable in a polynomial is called degree.
Zero degree polynomial: Constant polynomial, One term is constant e.g. 2, -5, 3 etc.
One degree polynomial: Linear polynomial, e.g. ax+b, where a, b, are constants, a
Two degree polynomial: Quadratic polynomial e.g. ax2+bx+c, where a, b, are constants, a
Three degree polynomial: Cubic polynomial, e.g. ax3+ bx+ cx+ d, where a, b, c, d, are constants, a
Not defined polynomial: Zero polynomial, Coefficient of each term is zero.
On the basis of Terms:
One term polynomial: Monomials e.g. 3, 4x etc.
Two terms polynomial: Binomials, e.g. ax + b, 2+ 5x, x-2y etc.
Three terms polynomials: Trinomials, e.g. ax2+ bx + c etc.
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ZEROES OF POLYNOMIALS
Zeroes (Roots) of a polynomial can be obtained by:
- Hit and Trial Method
- Solving the Polynomial Equation
HIT AND TRIAL METHOD: By putting x=a in a given polynomial and check
- If p(a) = 0, then ‘a’ is a zero of the given polynomial.
- If p(a) , then ‘a’ is not a zero of the given polynomial.
Example: Verify whether 3 and 0 are zeroes of the polynomial 2x2– 3x.
Let p(x)= 2x2-3x
Then p(3)= 2 32-3 = 18-9-9
p(0)= 0-0=0
Hence, 0 is a zero of the polynomial 2x2-3x, but 3 is not.
DOWNLOAD MOBILE APPLICATION TO LEARN MORE: POLYNOMIALS CLASS 9
DOWNLOAD MOBILE APPLICATION TO LEARN MORE: POLYNOMIALS CLASS 9
It focuses on
•Checking operation in division of numbers.
• Finding linear factor of p(x).
Example: Determine whether x-3 is a factor of polynomial
p(x) = x³-4x²-3x + 18
Sol. For x-3 to be a factor of p(x), by factor theorem, p(3) should be zero, where 3 is the zero of x-3
Now, by remainder theorem,
p(3)=3³-4.3²-3.3 + 18 =27-36-9 + 18
= 45-45 = 0
Hence, by Factor Theorem, x-3 is a factor of the given polynomial p(x).
FACTORISATION IN POLYNOMIALS
MONOMIALS: No factor
BINOMIALS: Two perfect squares and two perfect cubes.
TRINOMIALS: Quadratic polynomials p(x)= ax2+bx+c
1. By middle term spilliting.
p+q=b, pq = ac
(1) ax² + bx + c = x² + (p + q)x+ pq
(i) ax² + bx + c = (x + p) (x + q)
(iii) ax²-bx + c = (x-p)(x-q)
(iv) ax² bx- c = (x + p) (x – q)
Note: In point (iv).
- If b is negative, then negative sign goes with bigger number.
- If b is positive, then negative sign goes with smaller number.
2. If quadratic polynomial is a perfect square, then use
x² + 2xy + y² = (x + y)²
x² – 2xy + y² = (x – y)²
3. By using factor theorem.
CUBIC POLYNOMIALS: p(x) = ax + bx² + cx + d
- It has one real or distinct real root.
- One pair of repeated roots and a distinct roots.
- Can be found by using the following steps:
- Write all factors of constant term
- By trial, find p(q) = 0 . (x-q) is a factor of p(x)
- Divide p(x) by (x – q)
- Quotient is a quadratic equation.
- Factorise quadratic equation.
- ax³ + bx² + cx + d = (x-p)(x-q)(x-r)
ALGEBRAIC IDENTITIES
Identity I: (x + y)² = x² + 2xy + y²
Identity II: (x – y)² = x² – 2xy + y²
Identity III: x² – y² = (x + y)(x – y)
Identity IV: (x + a)(x + b) = x² + (a + b)x + ab
Identity V: (x+y+z)² = x² + y² + 2² + 2xy + 2yz + 2zx
Identity VI: (x + y)³ = x³ + y³ + 3xy(x + y)
Identity VII: (x − y)³ = x³ – y³ – 3xy(x – y)
= x³ – 3x²y + 3xy² – y³
Identity VIII: x³ +y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
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Table of Contents
POLYNOMIALS CLASS 9 CBSE -MATHEMATICS
1. The polynomial 2x – x2 + 5 is
(a) An equation
(b) A trinomial
(c) A binomial
(d) A monomial
Sol. B
2. Degree of a zero Polynomial is
(a) 0
(b) 1
(c) Any natural number
(d) Not defined
Sol. D, because zero polynomial means coefficient of any variable, i.e. 0x2, 0x3, 0x4, 0x5, …..is zero
3. Write True or False and justify your answer. The degree of the sum of two polynomials each of degree 5 is always 5.
Sol. False, because x5 + 1 and –x5 + 3x are two polynomials each of degree 5, but the degree of the sum of the polynomials is 1.
4. Write the degree of each of the following polynomials:
(i) 5x3 + 4x2 + 7x
(ii) 4 – y2
(iii) 5t –
(iv) 3
Sol. Degree of the given polynomials are
(i) 3
(ii) 2
(iii) 1
(iv) 3
5. Given a polynomial p(t) = t4-t³ +2²2 + 6, then p(-1) is
(a) 6
(b) 9
(c) 3
(d) – 1
Sol. p(t) = t4-t³ +2²2 + 6
⇒p(-1) = (-1)4 – (-1)³ + (−1)² + 6
= 1− (−1) + 1 +6=1+1+7= 9
Correct options is (b).
6. Zero of the zero polynomial is
(a) 0
(b) 1
(c) Any real number
(d) Not defined
Sol. (c) Consider g(x) = 0(x-a) where ‘a’ is any real number.
Zero of g(x) is equal to g(x) = 0⇒x-a=0
⇒x=a
So, zero of the zero polynomial is any real number.
7. If p(x)= x+3, then p(x) + p(-x) is equal to
(i) 3
(ii) 2x
(iii) 0
(iv) 6
Sol. p(x) + p(-x) = x + 3 + (-x) +3 = 6
Correct options is (d)
8. If p(x) = (x – 1) (x + 2), then we say,
(a) (x – 1) is a factor of p(x)
(b) (x + 1) is a factor of p(x)
(c) p(x) is divisible by both (x – 1) and (x + 2)
(d) All of these
Sol. d
9. Factors of 3x2 – x – 4 are
(a) (x – 1) and (3x – 4)
(b) (x + 1) and (3x – 4)
(c) (x + 1) and (3x + 4)
(d) (x – 1) and (3x + 4)
Sol.3x2 – x – 4 = 4x + 3x – 4
= x (3x – 4) + 1(3x – 4) = (x +1) (3x – 4)
Correct options: (b)
10. The expansion of (x – y) ³ is.
(a) x³ + y3 + 3x2y + 3xy2.
(b) x3 + y3 – 3x2y + 3xy2.
(c) x3 – y3 – 3x2y + 3xy²
(d) x3 – y3+ 3x2y– 3xy²
Sol. (c)
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