Pre-Secondary Scholarship Examination Grade 8 Maths Ch - Polynomials

Introduction to Algebraic Expression

  •  An algebraic expression is an expression made up of variables and constants along with mathematical operators.
  • A variable is a symbol, usually a letter, which is used to represent an unknown number. 
  • A term can be a number, a variable, or a number and variable combined by multiplication or division .In other words , a term is a product of variables and constants. A term can be an algebraic expression in itself.
  • An expression can be term or a collection of terms separated by addition or subtraction operators. 
  • The number (positive or negative) in the algebraic term is called the coefficients of the algebraic terms . The coefficient of 1 in an algebraic term is usually not written.
  • An algebraic expression can have any number of terms. The coefficient in each term can be any real number. There can be any number of variables in an algebraic expression. The exponent on the variables, however, must be rational numbers.
  • Example : 
 

What is a Polynomial ?

A Polynomial is an algebraic expression which includes constants, variables and exponents. It is the expression in which the variables have only positive integral powers.

Example

Note:

  • Polynomials are denoted by p(x), q(x) etc.

  • The constant term has no variable.

Examples of Polynomials:

3x2, 7xy +5, 6x2y-4xy3+(6/5)y.

Expressions which are not a Polynomials.

Polynomials in One Variable

An algebraic expression of the form a0 + a1x + a2x2+ .... + anxn where 

a0, a1, a2,....an are real numbers, n is a positive integer is called a polynomial in x.

If there is only one variable in the expression then this is called the polynomial in one variable.


You are familiar with factors and products in the case of numbers. For example, 8 is the product of 4 and 2. 4 and 2 are called the factors of 8. 

Similarly, the algebraic expression a. b. c = abc can be written as

  • 1.a.b.c or
  • 1.ab.c or
  • 1.bc. a or
  • 1.ac. b or
  • 1.abc.

1, a, b, c, ab, bc, ac, abc are all factors of a.b.c and a.b.c is a product.

 Degree of a Polynomial


  • For a polynomial in one variable – the highest exponent on the variable in a polynomial is the degree of the polynomial.
Example:
The degree of the polynomial x2-7x+19 is 2, as the highest power of x in the given expression is x2.
The degree of p(x) = x5 – x3 + 7 is 5.

Types Of Polynomials

Polynomials can be classified based on:
a) Number of terms
b) Degree of the polynomial.

Types of Polynomials on basis of number of Terms

(i) Constant polynomial:

  • A polynomial containing one term only, consisting of a constant is called a constant polynomial.
  • e.g., -6, 4, 95/60
  • Generally, each real number is a constant polynomial.

(ii) Zero polynomial:

  •  A polynomial consisting of one term, namely zero only, is called a zero polynomial.

(iii) Monomial:

  •  Polynomials having only one term are called monomials
    (‘mono’ means ‘one’).
  • e.g., a43 ,  (5/8)xz and – 2 are all monomials.

(iv) Binomial:

  •  Polynomials having only two terms are called binomials
    (‘bi’ means ‘two’).
  • e.g., (x2 + x), (y30 + √2) and (5x2y + 6xz) are all binomials.

(v) Trinomial:

  •  Polynomials having only three terms are calledtrinomials (‘tri’ means ‘three’).
  • e.g., (x^4 + x^3 + √2), (µ^43 + µ^7 + µ) and (8y – 5xy + 9xy^2 ) are all trinomials.

Types of Polynomials on basis of number of degrees

The highest value of the power of the variable in the polynomial is the degree of the polynomial.

a) Constant Polynomial 

  • A polynomial whose degree is zero is called a Constant Polynomial.

For example: 7/6 , -16 , 64 , √34 are constant polynomials.

b) Linear Polynomial

  • A polynomial whose degree is one is called a linear polynomial.

For example: 12x+1,  1/(2x – 7), √s + 5 are linear polynomials.

c) Quadratic Polynomial

  • A polynomial of degree two is called a quadratic polynomial.
  • In general, a quadratic polynomial can be expressed in the form ax2 + bx + c, where a≠0 and a, b, c are constants.

For example: 3x2+8x+5, x2 – 9, a2 + a + 7 are quadratic polynomials.

d) Cubic Polynomial

  • A polynomial of degree three is called a Cubic Polynomial.
  • In general, a quadratic polynomial can be expressed in the form ax3 + bx2 + cx + d, where a≠0 and a, b, c, d are constants.

For example, 3x3+8x+5 , x3 – 9x +2, a3 + a2 + √a + 7 are cubic polynomials.


 

 Value of Polynomial

Let p(y) is a polynomial in y and α could be any real number, then the value calculated after putting the value y = α in p(y) is the final value of p(y) at y = α. This shows that p(y) at y = α is represented by p (α).

Factorization

Factorization means expressing a given expression or number as a product of its factors. 

Factorization of Polynomials You know that any polynomial of the form p(a) can also be written as 

p(a)=g(a)*h(a)+R(a) i.e. Dividend = Quotient * Divisor + Remainder.

If the remainder is zero, then p(a) = g(a) x h(a). That is, the polynomial p(a) is a product of two other polynomials g(a) and h(a).

For example, 3a + 6a2 = 3a × (1 + 2a). 

It may be possible to express a polynomial as the product of two or more polynomials, in more than one form. 

the polynomial 3a + 6a2 = 3a × (1 + 2a) can also be factorised as

3a + 6a2 = 6a ×  (1/2 + a).

Methods of Factorizing Polynomials

There are various methods of factorizing a polynomial. They are,

  • 1. Factorization by dividing the expression by the HCF of the terms of the given expression.
  • 2. Factorization by grouping the terms of the expression.
  • 3.Factorization by splitting the middle term.
  • 4. Factorization using identities. 

Factorization by Dividing the Expression by the HCF of the Terms of the Given Expression

HCF of a polynomial is the largest monomial, which is a factor of each term of the polynomial. We can factorize a polynomial by finding the Highest Common Factor (HCF) of the terms of the expression and then dividing each term by its HCF. HCF and the quotient obtained are the factors of the given expression. 

Steps for Factorization

  •  Identify the HCF of the terms of the given expression.
  •  Divide each term of the given expression by the HCF and find the quotient.
  •  Write the given expression as a product of HCF and quotient.

Example :

If we have to factorize x2 –x then we can do it by taking x common.

x(x – 1) so that x and x-1 are the factors of x2 – x.

Factorization by Grouping the Terms of the Expression

In many situations, we come across polynomials, which may not have common factors among its terms. In such cases, we group the terms of the expression in such a way that there are common factors among the terms of the groups so formed. 

Steps for Factorization by Grouping

  •  Rearrange the terms if necessary.
  •  Group the given expression in such a way that each group has its common factor.
  •  Identify the HCF of each group. 
  • Identify the other factor.
  •  Write the expression as a product of the common factor and the other factor.

Example:

ab + bc + ax + cx = (ab + bc) + (ax + cx)

= b(a + c) + x(a + c)

= (a + c)(b + x)

Factorization of Trinomials of the form x2+bx+c

Trinomials are expressions with three terms.

For example, x2+ 14x + 49 is a trinomial.

There is no single method by which all trinomials can be factorized. We need to study the pattern in trinomials and choose the appropriate method to factorize the given trinomial.

Factorization of Trinomials by splitting the middle term

In these examples, study the relation between the middle and the last terms. 

Therefore, to factorize expressions of the type (x2 + bx +c), we have to find two factors which satisfy the above condition. That is, we need to split the middle term so that the product of the factors is equal to the last term.

Steps to Factorize a Trinomial of the form x2 + bx + c where b and c are Integers

  • Find all pairs of factors whose product is the last term of the trinomial.
  •  From the pairs of factors from step 1, choose a pair of factors whose sum is the coefficient of the middle term of the trinomial.
  • Split the middle term using the pair of factors from step 2 and rewrite the trinomial.
  • Group the terms from step 3 and factorize.
  • Verify the solution.
To factorize expressions of the type x2+ bx + c, you will find two numbers a and b such that their sum is equal to the coefficient of the middle term and their product is equal to the last term(constant).
  • Find the product (ac), of the coefficient of x2 and the last term.
  • List the factor pairs of ac.
  • Choose a factor pair whose sum is the coefficient of the middle term.
  • Rewrite the polynomial by splitting the middle term. 
  • Regroup and factorize.

x2 + bx + c = x2 + (p + q) + pq

= (x + p)(x + q)

This shows that we have to split the middle term in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.

 Example: 1

Factorize 6x2 + 17x + 5 by splitting the middle term.

Solution:

If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.

Some of the factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, out of which 2 and 15 is the pair which gives p + q = 17.

6x2 + 17x + 5 =6 x2 + (2 + 15) x + 5

= 6 x2+ 2x + 15x + 5

= 2 x (3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

Factorization using Identities Recall the following identities for finding the products:

Algebraic Identities
1. (x + y)2 = x2 + 2xy + y2
2. (x - y)2 = x2 - 2xy + y2
3. (x + y) (x - y) = x2 - y2
4. (x + a) (x + b) = x2 + (a + b)x + ab
5. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
6. (x + y)3 = x3 + y3 + 3xy(x + y) = x3+ y3 + 3x2y + 3xy2
7. (x - y)3 = x3- y3 - 3xy(x - y) = x3 - y3 - 3x2y + 3xy2
8. x3 + y3 = (x + y)(x2 – xy + y2)
9. x3 - y3 = (x - y)(x2 + xy + y2)
10. x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz - zx)
      x3 + y3 + z3 = 3xyz if x + y + z = 0 

Observe that the LHS in the identities are all factors and the RHS are their products. Thus, we can write the factors as follows:

  • Factors of a2 + 2ab + b2 are (a+b) and (a+b)

  • Factors of a2 _ 2ab + b2 are (a-b) and (a-b)

  •  Factors of a2 _ b2 are (a+b) and (a-b)

  •  Factors of  a3+ 3a2b + 3ab2 +b3 are (a+b), (a+b) and (a+b)

  •   Factors of  a3_ 3a2b + 3ab2 -b3 are (a-b), (a-b) and (a-b)

  • Factors of a2 + b2 + c2 +2ab+2bc+2ca  are (a+b+c ) and (a+b+c)
From the above identities we observe that a given expression, which is in the form of an identity can be written in terms of its factors.

Steps for factorization using Identities

  •  Recognize the appropriate identity.
  •  Rewrite the given expression in the form of the identity.
  •  Write the factors of the given expression using the identity.
Example:
 Factorize 8x3+ 27y3 + 36x2y + 54xy2

Solution:

The given expression can be written as

= (2x)3 + (3y)3 + 3(4x2) (3y) + 3(2x) (9y2)

= (2x)3+ (3y)3 + 3(2x)2(3y) + 3(2x)(3y)2

= (2x + 3y)3 (Using Identity VI)

= (2x + 3y) (2x + 3y) (2x + 3y) are the factors.
 

Example:

Factorize 4x2 + y2 + z2 – 4xy – 2yz + 4xz.
Solution:

The given expression can be written as
 
= (2x)2 + (–y)2 + (z)2 + 2(2x) (-y)+ 2(–y)(z) + 2(2x)(z)
 
= [2x + (- y) + z]2 (Using Identity V)
 

=(2x – y + z)2 = (2x – y + z) (2x – y + z)

Zeroes of a Polynomial

If p(x) is a polynomial then the number ‘a’ will be the zero of the polynomial with

p(a) = 0. We can find the zero of the polynomial by equating it to zero.

In other words ,

Zeroes of a polynomial p(x) is a number a such that p(a) = 0.

* Zero may be a zero of a polynomial.
* Every linear polynomial has one and only one zero.
* Zero of a polynomial is also called the root of the polynomial.
* A non-zero constant polynomial has no zero.
* Every real number is a zero of the zero polynomial.
* A polynomial can have more than one zero.

The maximum number of zeroes of a polynomial is equal to its degree.

 Example:

Given polynomial is p(x) = x - 4

To find the zero of the polynomial we will equate it to zero.

x - 4 = 0

x = 4

p(4) = x – 4 = 4 – 4 = 0

This shows that if we place 4 in place of x, we got the value of the polynomial as zero. So 4 is the zero of this polynomial. And also we are getting the value 4 by equating the polynomial by 0.

So 4 is the zero of the polynomial or root of the polynomial.

The root of the polynomial is basically the x-intercept of the polynomial.

 

If the polynomial has one root, it will intersect the x-axis at one point only and if it has two roots then it will intersect at two points and so on.

Relationship between Zeroes and Coefficients of a Polynomial

For Quadratic Polynomial: 

  • If α and β are the zeroes of the quadratic polynomial p(x) = ax2 + bx + c, a ≠ 0, then we know that (x – α) and (x – β) are the factors of p(x). 

  • α +  β = -b/a

    Sum of zeroes = -coefficient of x /coefficient of x2

  •  αβ = c/a

    Product of zeroes = constant term / coefficient of x2

For Cubic Polynomial

If α,β and γ are the roots of a cubic polynomial ax3+bx2+cx+d, then

α+β+γ = -b/a

αβ +βγ +γα = c/a

αβγ = -d/a

Division Algorithm for Polynomials

If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x).

To divide one polynomial by another, follow the steps given below.

Step 1: arrange the terms of the dividend and the divisor in the decreasing order of their degrees.

Step 2: To obtain the first term of the quotient, divide the highest degree term of the dividend by the highest degree term of the divisor  Then carry out the division process.

Step 3: The remainder from the previous division becomes the dividend for the next step. Repeat this process until the degree of the remainder is less than the degree of the divisor.


 One more Example :


Remainder Theorem

If f(x) is a polynomial in x and is divided by x-a; the remainder is the value of f(x) at x = a i.e., Remainder = f(a).

Proof: Let p(x) be a polynomial divided by (x-a). Let q(x) be the quotient and R be the remainder.

By division algorithm,

Dividend = (Divisor x quotient) + Remainder

p(x) = q(x) . (x-a) + R

Substitute x = a,

p(a) = q(a) (a-a) + R p(a) = R (a - a = 0, 0 - q (a) = 0)

Hence Remainder = p(a).

Steps for Factorization using Remainder Theorem

  •  By trial and error method, find the factor of the constant for which the given expression becomes equal to zero.
  •  Divide the expression by the factor that is determined in step 1.
  •  Factorize the quotient. If the quotient is a trinomial, factorize it further.
  •  If the expression is a 4th degree expression, the first step will be to reduce this to a trinomial and then factorize this trinomial further.

Division of a Polynomial with a Monomial

 

We can see that ‘x’ is common in the above polynomial, so we can write it as

 

Hence 3x^2 + x + 1 and x the factors of 3x^3 + x^2 + x.

Steps of the Division of a Polynomial with a Non –Zero Polynomial

Divide x^2 - 3x -10 by 2 + x

Step 1:  Write the dividend and divisor in the descending order i.e. in the standard form. x^2 - 3x -10 and x + 2

Divide the first term of the dividend with the first term of the divisor.

x^2/x = x this will be the first term of the quotient.

Step 2: Now multiply the divisor with this term of the quotient and subtract it from the dividend.

 

Step 3: Now the remainder is our new dividend so we will repeat the process again by dividing the dividend with the divisor.

Step 4: – (5x/x) = – 5 

Step 5: 

 

The remainder is zero.

Hence x^2 - 3x – 10 = (x + 2)(x - 5) + 0

Dividend = (Divisor × Quotient) + Remainder

So,we can conclude by saying,

Remainder Theorem says that if p(x) is any polynomial of degree greater than or equal to one and let ‘t’ be any real number and p (x) is divided by the linear polynomial x – t, then the remainder is p(t).

As we know that

P(x) = g(x) q(x) + r(x)

If p(x) is divided by (x-t) then

If x = t

P (t) = (t - t)*q (t) + r = 0

To find the remainder or to check the multiple of the polynomial we can use the remainder theorem.

Example:

What is the remainder if a4 + a3 – 2a2 + a + 1 is divided by a – 1.

Solution:

P(x) = a^4 + a^3 – 2a^2 + a + 1

To find the zero of the (a – 1) we need to equate it to zero.

a -1 = 0

a = 1

p (1) = (1)^4 + (1)^3 – 2(1)^2 + (1) + 1

= 1 + 1 – 2 + 1 + 1

= 2

So by using the remainder theorem, we can easily find the remainder after the division of polynomial.

Factor Theorem

 If p(x), a polynomial in x is divided by x-a and the remainder = p (a) is zero, then (x-a) is a factor of p(x). 

Proof: When p(x) is divided by x-a,

R = p(a) (by remainder theorem)

p(x) = (x-a).q(x)+p(a) (Dividend = Divisor x quotient + Remainder Division Algorithm)

But p(a) = 0 is given.

Hence p(x) = (x-a).q(x) 

Conversely if x-a is a factor of p(x) then p(a) = 0.

p(x) = (x-a).q(x) + R

If (x-a) is a factor, then the remainder should be zero (x - a divides p(x) exactly)

R = 0 By remainder theorem,

but ,R = p(a) 

thus, p(a) = 0

So,we can conclude by saying,

Factor theorem says that if p(y) is a polynomial with degree n≥1 and t is a real number, then

  1. (y - t) is a factor of p(y), if p(t) = 0, and

  1. P (t) = 0 if (y – t) is a factor of p(y).

Example:

Algebraic Identities

1. (a+b)2=a2+2ab+b2
2. (a−b)2=a2−2ab+b2
3. (x+a)(x+b)=x2+(a+b)x+ab
4. a2−b2=(a+b)(a−b)
5. a3−b3=(a−b)(a2+ab+b2)
6. a3+b3=(a+b)(a2−ab+b2)
7. (a+b)3=a3+3a2b+3ab2+b3
8. (a−b)3=a3−3a2b+3ab2−b3

 Note : 

1) the degree of zero is not defined. 

2)The method of addition subtraction multiplication and division is similar to the operations of algebraic expressions for polynomials. 

3) The degree of the product of two polynomials is equal to the degree of multiplicand + the degree of multiplier .

example-  the degree of the product of x^4 + 5 and X^2 + 3 is 4 + 2 is equal to 6.

4)