Question

over all the importance concept in lesson polynomial class 9 with example for each
answer should be in brief

Answer 1

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Summary of
☆POLYNOMIALS CLASS 9 CBSE ☆

> Polynomials in one variable : Expressions with whole numbers as exponents of variable

e.g. : 4x^2 - 3x + 7 , because in this, the only variable x is having a whole number exponent, that is, 2

Terms :
In the expression 3q+ t, both 3q and t are different terms.

> Coeffecient : The numerical value accompanied with a variable.

e.g. : in 2x, 2 is the numerical Coeffecient of x

> The constant polynomial 0 is called Zero polynomial

> On the basis of number of terms, algebraic expressions are :

• BINOMIAL : 2 terms
e.g. : x+2

• TRINOMIAL : 3 terms
e.g. : 4+ 3x - 6

• POLYNOMIAL : More than 1 term.

> Degree : The highest power of an expression.

e.g. : In 3q+t^3 + 8, 3 is degree.

* The degree of non-constant polynomial is 0

e.g. : Degree of 5 is 0

* Degree of 0 isn't defined.

> On the basis of degree, the polynomials are divided :

• A polynomial with degree one is Linear polynomial
e.g. : 2x-1

• A polynomial with degree 2 is Quadratic Polynomial
e.g. : 5-y^2

•A polynomial with degree 3 is called a cubic polynomial
e.g. : x^2 - 3 + x^3

> A linear polynomial has only one and only ONE zero.

*A Zero of the polynomials need not be 0
* 0 may be a 0 of polynomial.
* A polynomial can have more than one zero.

> *Remainder Theorom*

[ The general form doesn't make any one understand ;p. so, just wish to explain by an example ]

Q : Find remainder when x^4 + x^3 - 2x^2 + x + 1 is divided by x-1

•First , find zero
here, x-1
x = 1

• Substitute the zero in equation
P (1) = (1)^4 + (1)^3 - 2×(1)^2 + 1 + 1
= 1+1-2+1+1 = 2

• That's all about remainder theorom!

> Long division method can't be explained here. so, pls visit : veedyavision institute.com for the method (YouTube)

> Btw, i can give you general form of Remainder Theorom ;p

* If p (x) is divided by x-a, then p (a) is remainder.

> Factorization (can be defined as The most worse part ;p)

> Factorisation can be done by two methods :

1. Splitting the middle term
2. Factor theorom

1. SPLITTING THE MIDDLE TERM

[ by an example ;p]

Q. Factorise 6x^2 + 17x + 5

• Multiply first and last numericals (pq)
here, 6×5 = 30

• p+q = the middle term
here, p+q = 17

• Look at some factors of pq,
here, factors of 30

1. 30 = 2×15
2. 30 = 3×10
3. 30 = 5×6

•Choose the factors which gives p+q
here,
1 is correct to choose as 2+ 15 gives p+q = 17 (15+2)

• Split the middle term!
here, 6x^2 + 17x + 5
= *6x^2 + 2x* + *15x + 5* (The grp in closed within ** is one grp ;p)

[ 17x split into 15x and 2x ]

• Take common terms out!

here, From grp 6x^2 + 2x (2x common)
2x (3x + 1)

From grp, 15x + 5 (5 common)
5 (3x + 1)

• Take the common as one and seperated as other

here, (2x + 5) (3x+1) [Answer]

2. Factor theorom

[ general form : x-a is a factor of polynomial p (x), if p (a) = 0. Also, if x-a is a factor of p (x), then p (a) = 0 ]

• Take the factors of last term, (commonly a numerical)

• Trail it to know which are the zeros of the given polynomail

• if a is a zero, x-a is a factor

[An important thing to remember : The degree of the polynomial = No.of factors]

> Algebraic Expressions

> Basic Identities (we have studies so far)

> Advanced ones

• (x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx

• (x+y)^3 = x^3 + y^3 + 3xy (x+y)
or, x^3 + 3x^2y + 3xy^2 + y^2

• (x-y)^3 = x^3 - y^3 - 3xy (x-y)

• x^3 + y^3 + z^3 - 3xyz = (x+y+z) (x^2 + y^2 + z^2 - xy - yz - zx)

Hope it helps.♥
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Have more doubts? ping me any time ;p