Class IX Unit-2 Polynomials

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Constants and Variable                                                            

The value of constant remains same throughout the case and it does not changes in the given situation. They are generally denoted by a,b,c etc
Variable value keep changing and they are generally denoted by the letters x,y,z etc

What is a Polynomial expression

A polynomial expression S(x) in one variable x is an algebraic expression in x term as
S(x)=a_nxⁿ+a_(n-1)x^(n-1)+a_(n-2)x^(n-2)+....+ax+a₀
Where a_n,a_n-1,...,a,a₀ are constant and real numbers and an is not equal to zero

Some important points to remember
1) a_n,a_n-1,...,a,a₀ are called the coefficients for xⁿ,x^n-1 ,..,x,x⁰
2) n is called the degree of the polynomial
3) when a_n,a_n-1,....,a,a₀ it is called zero polynomial
4) A constant polynomial is the polynomial with zero degree, it is a constant value polynomial
5) A polynomial of one item is called monomial, two items binomial and three items as trinomial
6) A polynomial of one degree is called linear polynomial, two degree as quadratic polynomial and degree three as cubic polynomial

We can also have polynomials in more than one variable. For example, x² + y² +z² (where variables are x, y and z) is a polynomial in three variables. Similarly p² + q⁴ + r (where the variables are p, q and r) is polynomials in three variables

Also if a₀ = a₁ = a₂ = a₃ = . . . = a_n = 0 (all the constants are zero), we get the zero polynomial, which is denoted by 0. Now we may ask What is the degree of the zero polynomial? The degree of the zero polynomial is not defined

how to find the degree of a polynomial

The degree of a polynomial is the greatest exponent of the variable in the polynomial when the polyomial is expression in its canonical form consisting of a linear combination of monomials. 
The degree of a term is the sum of the exponents of the variables that appear in it

Lets take a example of polynomial
S(x) =x³ +x²+ 5
Here the highest exponent of x is 3. So degree of the polynomail is 3

Similarly

Polynomial	Degree of the polynomial
4x +1	1
3x2 + 4x+6	2
9x3 -4x2 +1	3
x5 -x+1	5

We have polynomial types depending on degree also

Polynomial	Degree of the polynomial	Polynomial type
4x +1	1	Linear polynomial
3x2 + 4x+6	2	Quadratic Polynomial
9x3 -4x2 +1	3	Cubic Polynomial
x4 -x+1	4	Quartic Polynomial
5x5 -x+1	5	Quintic Polynomial

For constant polynomial
P(x) = 9
It can be expression as
P(x) =9x⁰

So degree is zero

Value of the polynomial:

Lets take a example of polynomial
S(x) =x² +1
Then
When we put the value of x=2,then
S(2)=4+1=5
The S(2) is the called the value of polynomial at x=2
In General terms, the value of polynomial at x=a is S(a)

Zero's or roots of the polynomial

It is a solution to the polynomial equation S(x)=0 i.e. a number "a" is said to be a zero of a polynomial if S(a) = 0.
If we draw the graph of S(x) =0, the values where the curve cuts the X-axis are called Zeros of the polynomial
a) Linear polynomial has only one root
b) A zero polynomial has all the real number as roots
c) A constant polynomial has no zeros
d) A zero of polynomial need not to be 0 

Adding polynomials

We many times need to add the two polynomial .Adding polynomial just means adding the like terms.We need to follow below steps for Addition of polynomial 
1) Arrange both the polynomial is same order of exponent . It would be good to have terms arrange from highest exponent to lowest exponent i.e
For example 
S(x) = x² + 5x³ + 1 +10x
P(x) = x³ + x² + 5 +10x 

Arranging them as per suggestion above

S(x) = 5x³ +x²+ 10x +1
P(x) = x³ + x² + 10x +5 

2) Add the like term . By like term ,we mean same exponent terms. There are two method to add like terms. We either can group them horizontally and add it. We can add vertically 
Horizontally
S(x) + P(x)
=x² + 5x³ + 1 +10x + (x³ + x² + 5 +10x)
Arranging both the polynomial in same order
=5x³ +x²+ 10x +1 + (x³ + x² + 10x +5 )
Opening the parentthesis and grouping the like terms. In addition,no sign are changed when parentthesis are opened
= 5x³ + x³ + x²+x² + 10x +10x + 1+5
Adding the like term 
=6x³+2x²+20x+6

Vertically
S(x) = x² + 5x³ + 1 +10x
P(x) = x³ + x² + 5 +10x 

Arranging both the polynomial in same order

S(x) = 5x³ +x²+ 10x +1
P(x) = x³ + x² + 10x +5 



Similarly we can add three or more polynomials

Example-1
(2x³ - x+1+x²) + (x³ + 6x - 7) + (-3x² - 11 + 2x)
Arranging them in same order
=(2x³ + x² - x+1) + (x³ + 6x - 7) + (-3x² + 2x - 11)

Opening the parentthesis and grouping the like terms
=2x³+x³ +x²-3x² -x +6x+2x +1-7-11
=3x³-2x²+7x-16

Subtracting polynomials

We many times need to subtract the two polynomial .It is very similar to adding polynomails only.Subtracting polynomials just means Subtracting the like terms.We need to follow below steps for Subtraction of polynomial 
1) Arrange both the polynomial is same order of exponent . It would be good to have terms arrange from highest exponent to lowest exponent i.e
For example 
S(x) = x² + 5x³ + 1 +10x
P(x) = x³ + x² + 5 +10x 

Arranging them as per suggestion above

S(x) = 5x³ +x²+ 10x +1
P(x) = x³ + x² + 10x +5 

2) Subtract the like term . By like term ,we mean same exponent terms. There are two method to subtract like terms. We either can group them horizontally and subtract it. We can add vertically 
Horizontally
S(x) - P(x)
=x² + 5x³ + 1 +10x - (x³ + x² + 5 +10x)
Arranging both the polynomial in same order
=5x³ +x²+ 10x +1 -(x³ + x² + 10x +5 )
Opening the parentthesis and grouping the like terms. In Subtract, sign are reversed when parentthesis are opened i.e + becomes - and - becomes +
= 5x³ - x³ + x² - x² + 10x -10x + 1-5
Subtact the like term 
=4x³ - 4 

Vertically
S(x) = x² + 5x³ + 1 +10x
P(x) = x³ + x² + 5 +10x 

Arranging both the polynomial in same order

S(x) = 5x³ +x²+ 10x +1
P(x) = x³ + x² + 10x +5 



Similarly we can add three or more polynomials

Example-1
(2x³ - x+1+x²) - (x³ + 6x - 7) - (-3x² - 11 + 2x)
Arranging them in same order
=(2x³ + x² - x+1) - (x³ + 6x - 7) - (-3x² + 2x - 11)

Opening the parentthesis and grouping the like terms
=2x³- x³ +x²+3x² -x -6x-2x +1+7+11
=x³+4x²-9x+18

multiplying polynomials

Multiplication of polynomial is simple task and we can follow below steps to multiply polynomials

1)Arrange both the polynomial is same order of exponent . It would be good to have terms arrange from highest exponent to lowest exponent i.e
2) We have to distribute each term of the first polynomial to every term of the second polynomial. 
a) when you multiply two terms together you must multiply the coefficient (numbers) and add the exponents
b) Also as we already know ++ equals =, +- or -+ equals - and -- equals +

So multiplication may involve 
a) Multiplication of monomial to monomial
b) Multiplication of monomial to binomial, trinomial or more terms polynomials
c) Multiplication of binomial, trinomial or more terms polynomials to monomial
d) Multiplication of binomial to binomial, trinomial or more terms polynomials
e) Multiplication of trinomial to trinomial or more terms polynomials

Multiple the Monomials

1. x² x (2x²²) x (4x²⁶)

      2 (-10xy³/3) ×(6x^3y/5)

      3.    ( x) x (x²) x (x³) x (x⁸)

Answer: 

We will use the below property extensively in above questions

x^m x xⁿ x x^o = x^m+n+o

1. As you know

           So, we get

       a² x (2a²²) x (4a²⁶) = 8a⁴⁸

2. (2xy/3) ×(-9x²y²/10)

=(-3x³y³/5)

3.  ( x) x (x²) x (x³) x (x⁸)

= x¹⁴

Multiply the binomials.

i) (2x + 5) and (4x – 3)

 (ii) (x – 8) and (3x – 4)

 (iii) (2.5x – 0.5) and (2.5x + 0.5)

Answer: 

Let  ( a+b) (c+d) to be done

then

( a+b) (c+d)= a(c+d) + b( c+d)

=(a x  c)+(a x  d)+(b x  c)+(b x d)

We will use the same concept in all  the question below

1) (2x + 5)(4x - 3)

= 2x x 4x - 2x x 3 + 5 x 4x - 5 x 3

= 8x² - 6x + 20x -15

= 8x² + 14x -15

2) ( y - 8)(3y - 4)

= y x 3y - 4y - 8 x 3y + 32

= 3y² - 4y - 24y + 32

= 3y² - 28y + 32

3) (2.5l - 0.5m)(2.5l + 0.5)

Using (a+b)(a-b) = a² - b²

We get = 6.25l² - 0.25m²

Multiply Binomial by Trinomial

(3x + 2)(4x² – 7x + 5)

Answer

=12x³ -21x² +15x +8x² -14x+10

=12x³ -13x² +x+10

Dividing Polynomails

When a polynomial p(x) is divided by the polynomial g(x), we get quotient q(x) and remainder r(x)

p(x)=g(x).q(x)+r(x) 
Notes
1) The degree of the reminder r(x) is always less then divisor g(x)

Now Let us see how to divide the polynomial by another non-zero polynomial

Steps to divide a polynomial by another polynomial. This is also called the long division method of polynomials

1) Arrange the term in decreasing order in both the polynomial
2) Divide the highest degree term of the dividend by the highest degree term of the divisor to obtain the first term
3) Now We multiply the divisor by the first term of the quotient, and subtract this product from the dividend
3) Similar steps are followed till we get the reminder whose degree is less than of divisor

Example Divide p(x) by g(x), where p(x) = x + 4x² -1 and g(x) = 1 + x.
Solution
We carry out the process of division by means of the following steps:
Step 1 : We write the dividend x + 4x² - 1 and the divisor 1 + x. in the standard form, i.e., after arranging the terms in the descending order of their degrees. So, the dividend is 4x² + x �1 and divisor is x + 1.
Step 2 :We divide the first term of the dividend by the first term of the divisor, i.e., we divide 4x² by x, and get 4x. This gives us the first term of the quotient.
Step 3 :We multiply the divisor by the first term of the quotient, and subtract this product from the dividend, i.e., we multiply x + 1 by 3x and subtract the product 4x² + 4x from the dividend 4x² + x - 1. This gives us the remainder as -3x - 1.
Step 4 : We treat the remainder -3x - 1 as the new dividend. The divisor remains the same. We repeat Step 2 to get the next term of the quotient, i.e., we divide the first term -3x of the (new) dividend by the first term x of the divisor and obtain �3. Thus, �3 is the second term in the quotient.
Step 5 We multiply the divisor by the second term of the quotient and subtract the product from the dividend. That is, we multiply x + 1 by - 3 and subtract the product -3x - 3 from the dividend - 3x - 1. This gives us 2 as the remainder

Now the degree of the reminder is less than degree of the divisor so process stops here

Remainder Theorem's

If p(x) is an polynomial of degree greater than or equal to 1 and p(x) is divided by the expression (x-a),then the reminder will be p(a)

Important notes 
1) for (x-a) then remainder P(a)
2) for (x+a) => x -(-a),then remainder will be P(-a)
3) for (ax-b) => a(x-b/a) ,the remainder will be P(b/a)
4) for (ax+b) => a(x+b/a),the remainder will be P(-b/a)
5) for (b-ax)=> -a(x-b/a),the remainder will be P(b/a) 

How to factor polynomials

We have studied about polynomial addition,subtraction ,multiplication and division. Now it time to learn How to factor polynomials. As you are already aware factoring is the process by which we go about determining what we multiplied to get the given quantity. So factoring polynomials means finding the terms which when multiplied together produced the polynomials
The various methods to perform factoring of polynomails are

Greatest Common Factor

We can look at each of the term in the polynomials ,factorize each term and then find common factors to factorize the expression. This should be the first steps always when you are factoring the polynomials

In these problem, we need to factorize each term and then find common factors to factorize the expression

i) 7x-42

=  (7×x) – (7×6)

=7(x-6)

ii) 6p-12

=  (6×p) – (6×2)

=6(p-2)

iii) 7a²+14a

=(7×a×a) +  (2×7×a)

=7a(a+2)

iv) -16z+ 20z³

=4z(5z²-4)

v) 20 l² m + 30 a l m

=(2×2×5×l×m×l)  + (2×3×5×a×l×m)

=2lm(10l+15a)

vi) 5 x²y -15 xy²

=(5×x×x×y)  + (5×3×x×y×y)

=5x(xy-3y²)

vii) 10a² -15 b²+20c²

=(2×5×a×a)  - (5×3×b×b) + (5×2×2×c×c)

=5(2a²-3b²+4c²)

viii) -4a² +4a b-4ca

=(-4×a×a)  +(4×b×a) - (4×c×a)

=-4a(a-b+c)

ix)  x²yz + xy²z  +xyz²

=(x×x×y×z)  +(x×y×y×z) +(x×y×z×z)

=xyz(x+y+z)

x) ax²yz + bxy²z  +cxyz²

=(x×x×y×z×a)  +(x×y×y×z×b) +(x×y×z×z×c)

=xyz(ax+by+cz)

factor polynomials by grouping

When we dont see common factor across all the terms, we may look at grouping the terms and check if we find binomial factor from both the groups .Let us take example to take close look at the method

1) x² + 8x - x - 8
Step 1 Check if there is a common factor among all terms. There is none.
Step 2 Think of grouping. Notice that first two terms have a common factor x;
x² + 8x = x(x+8)
What about the last two terms? Observe them. If you take -1 out
-x-8 = -1(x+8)
Step 3 Putting both part togethes together,
x² + 8x - x - 8= x(x+8) -1(x+8)
= (x+8)(x-1)
2) x⁵ -2x³ -2x² +4
Step 1 Check if there is a common factor among all terms. There is none.
Step 2 Think of grouping. Notice that first two terms have a common factor x;
x⁵ -2x³ = x³(x²-2)
What about the last two terms? Observe them. If you take -2 out
-2x² +4 = -2(x²-2)
Step 3 Putting both part togethes together,
x⁵ -2x³ -2x² +4= x³(x²-2) -2(x²-2)
= (x²-2)(x³-2)

factor Quadratic polynomials by splitting the middle term

Let the Quadratic polynomial be
P(x) =ax² +bx +c
Let its factors be (px + q) and (rx + s). Then
ax² +bx +c= (px + q) (rx + s) = pr x² + (ps + qr) x + qs
Comparing the coefficients of x², 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 ax² +bx +c, we have to write b as the sum of two numbers whose product is ac

1) 4x² + 10x - 6
Step 1 Here b=10 and ac=-24
Step 2 We need to find factor of -24,whose sum is 10, the number which satisfy this is 12 and -2
4x² + 10x - 6
=4x² +(-2+12)x -6
=4x² -2x + 12x -6
= 2x(2x-1)+ 6(2x-1)
= (2x-1)(2x+6)
2) x⁵ -2x³ -2x² +4

Step 1 Check if there is a common factor among all terms. There is none.
Step 2 Think of grouping. Notice that first two terms have a common factor x;
x⁵ -2x³ = x³(x²-2)
What about the last two terms? Observe them. If you take -2 out
-2x² +4 = -2(x²-2)
Step 3 Putting both part togethes together,
x⁵ -2x³ -2x² +4= x³(x²-2) -2(x²-2)
= (x²-2)(x³-2)

Factor's Theorem's

If x-a is a factor of polynomial p(x) then p(a)=0 or if p(a) =0,x-a is the factor the polynomial p(x) 

4) Factorising a Polynomial by Factor Theorem

We know by factor theorem if (x-a) is the factor of the polynomial ,then P(a)=0. We can use this theorem to find factors of the polynomials

Lets first take a look at quadratic polynomial
1) 4x² + 10x - 6
Step 1 Take 4 out of the expression 4( x² + 5x/2 -3/2)
Step 2 This can be further expressed as 
4( x² + 5x/2 -3/2) = 4(x-a) (x-b)
Here so ab= -3/2
We need to look for all the factor of the term -3/2 i.e +1/2,-1/2,-3,+3,1/4,-1/4,3/2,-3/2
Whereever it satisfies the factor theorem, we are good
In this particular case
P(1/2)=P(-3)=0, we can write like this

4( x² + 5x/2 -3/2) = 4(x-1/2) (x+3)
= (2x-1)(2x+6)
For the last example, it may seems splitting the terms seems more efficient
Now let us take a look at cubic polynomial
Suppose the Polynomial is the form
P(x)= x³ +6x²+11x+6

Step 1 We need to look at the constant 6 and factorise it
The factor of 6 will be 1,2,3
Now we can try the polynomial for all the values -3,-2,-1,1,2,3
Whereever it satisfies the factor theorem, we are good
In this particular case
P(-1)=P(-2)=P(-3)=0, we can write like this
Step 2 x³ +6x²+11x+6=K(x+1)(x+2)(x+3)

We can put any value of x in this identity and get the value of x
In this particular case putting x=0, we get K=1

So the final identity becomes
x³ +6x²+11x+6=(x+1)(x+2)(x+3)

Or we can just find the one factor (x+1) and try the splitting or long division method to get the remaining quadratic polynomial and solve that by splitting method
x³ +6x²+11x+6
= x³ +x²+5x²+5x+6x +6
Splitting cubic polynomial to get (x+1)
=x²(x+1) + 5x(x+1) + 6(x+1)
=(x+1)(x²+5x+6)
Now solving the quadratic polynomial by splitting method
=(x+1)(x²+2x+3x+6)
=(x+1)[x(x+2)+3(x+2)]
=(x+1)(x+2)(X+3)
In General Term,

S(x)=a_nxⁿ+a_(n-1)x^(n-1)+a_(n-2)x^(n-2)+....+ax+a₀

Look for the factors in a₀/a_n, Take both the positive and negative values and find out which suites your polynomial and then find the value of k

Factoring polynomial using 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 + z² + 2xy + 2yz + 2zx

Identity VI : (x + y)³ = x³ + y³ + 3xy (x + y)

Identity VII :(x - y)³ = x³ - y³ - 3xy(x - y)

Identity VIII : x³ + y³ + z³ -3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

Identity IX : x³ + y³ = (x + y )(x² + y² - xy )

Identity X : x³ - y³ = (x - y )(x² + y² + xy )

1) 27x³ +1
Step 1 Both th terms are exact cubes (3x) and 1
Step 2 We can the algebric identity IX
27x³ +1
=(3x+1)(9x²+1-3x)

2)8x³ + y³ + 27z³ -18xyz
= (2x)³ + y³ + (3z)³ - 3(2x)(y)(3z)
= (2x + y + 3z)[(2x)² + y² + (3z)² - (2x)(y) - (y)(3z) - (2x)(3z)]
=
(2x + y + 3z) (4x² + y² + 9z² - 2xy - 3yz - 6xz)

Solved Examples Polynomials

1) If x + 1/x = 11, evaluate x² + 1/x²

Solution x + 1/x = 12
(x + 1/x)² = (x)² +1/x² + 2(x)(1/x) = 144
x² + 1/x² + 2 = 144
x²+ 1/x² = 144 - 2 = 142

2) If 9x²+ 25y² = 181 and xy = -6. Find the value of 3x + 5y

Solution 9x²+ 25y² = 181
(3x)² + (5y)² = 181
(3x)² + (5y)² + 30xy - 30xy = 181
(3x)² + (5y)² + 2(5x)(6y) = 181 + 30xy
(3x + y)² = 181 + 30(-6) = 181 - 180
(3x + y)² =1
3x + y = +1 or -1

3) Prove that x² + y² + x² -xy -yz -zx is always non-negative for all values of x, y and y.

Solution To prove x² + y² + x² -xy -yz -zx = 0.
We know that square of any number (+ve or -ve) is always +ve.
x² + y² + z² -xy -yz -zx
= (1/2)[2x² + 2y² + 2z² -2xy -2yz -2zx]
=(1/2)[x² + y² -2xy + y² + z² -2yz + x² + z² -2zx]
=(1/2)[(x² + y² -2ab) + (y² + z² -2yz) + (x² + z² -2zx)]
=(1/2)[(x - y)² + (y - z)² + (z - x)² ]
Here, all the terms are always be positive,
x² + y² + x² -xy -yz -zx = 0