Polynomials

Hello my dear students! Welcome to today's mathematics lesson. I am so happy to see you all here, ready to learn something new and exciting. Today, we are going to study Chapter 2 of your NCERT mathematics textbook, and the name of this chapter is "Polynomials". This is a very important chapter that builds upon what you have learned in earlier classes about algebraic expressions. So let's begin our journey into the world of polynomials together.

In the beginning of this chapter, you will recall that you have already studied algebraic expressions, their addition, subtraction, multiplication, and division in your earlier classes. You have also learned how to factorise some algebraic expressions using certain algebraic identities. Do you remember the identities we studied? Let me remind you:

We have (x + y)² = x² + 2xy + y², which is the square of the sum of two terms.

We have (x - y)² = x² - 2xy + y², which is the square of the difference of two terms.

And we have x² - y² = (x + y)(x - y), which is the difference of squares.

These identities are very useful in factorisation, and we will be using them extensively in this chapter. In this chapter, we shall start our study with a particular type of algebraic expression called a polynomial, and we will learn the terminology related to it. We will also study the Remainder Theorem and the Factor Theorem, and their use in the factorisation of polynomials. In addition to this, we will study some more algebraic identities and their use in factorisation and in evaluating some given expressions.

Now, let us begin with Section 2.2, which is titled "Polynomials in One Variable".

Let us first recall what a variable is. A variable is denoted by a symbol that can take any real value. We use the letters x, y, z, and so on to denote variables. Now, notice that 2x, 3x, -x, and -½x are all algebraic expressions. All these expressions are of the form (a constant) × (a variable). Now suppose we want to write an expression which is a constant times a variable, and we do not know what the constant is. In such cases, we write the constant as a, b, c, and so on. So the expression will be ax, for instance.

However, there is a very important difference between a letter denoting a constant and a letter denoting a variable. The values of the constants remain the same throughout a particular situation, that is, the values of the constants do not change in a given problem, but the value of a variable can keep changing. This is a fundamental distinction that you must remember.

Now, let us consider a practical example. Consider a square of side 3 units. What is its perimeter? You know that the perimeter of a square is the sum of the lengths of its four sides. Here, each side is 3 units. So, its perimeter is 4 × 3, that is, 12 units. What will be the perimeter if each side of the square is 10 units? The perimeter is 4 × 10, that is, 40 units. Now, what if the length of each side is x units? Then the perimeter is given by 4x units. So, as the length of the side varies, the perimeter varies. This is a perfect example of how a variable works.

Now, can you find the area of the square? It is x × x = x² square units. x² is an algebraic expression. You are also familiar with other algebraic expressions like 2x, x² + 2x, x³ - x² + 4x + 7. Now, here is something very important to notice: all the algebraic expressions we have considered so far have only whole numbers as the exponents of the variable. Expressions of this form are called polynomials in one variable. In the examples above, the variable is x. For instance, x³ - x² + 4x + 7 is a polynomial in x. Similarly, 3y² + 5y is a polynomial in the variable y, and t² + 4 is a polynomial in the variable t.

Now, let us understand the structure of a polynomial. In the polynomial x² + 2x, the expressions x² and 2x are called the terms of the polynomial. Similarly, the polynomial 3y² + 5y + 7 has three terms, namely, 3y², 5y, and 7. Can you write the terms of the polynomial -x³ + 4x² + 7x - 2? This polynomial has 4 terms, namely, -x³, 4x², 7x, and -2.

Each term of a polynomial has a coefficient. So, in -x³ + 4x² + 7x - 2, the coefficient of x³ is -1, the coefficient of x² is 4, the coefficient of x is 7, and -2 is the coefficient of x⁰. Remember, x⁰ = 1. So, -2 is actually the constant term. Do you know the coefficient of x in x² - x + 7? It is -1. Let me explain: the term -x can be written as -1 × x, so the coefficient of x is -1.

Now, 2 is also a polynomial. In fact, 2, -5, 7, and so on are examples of constant polynomials. The constant polynomial 0 is called the zero polynomial. This plays a very important role in the collection of all polynomials, as you will see in the higher classes.

Now, consider algebraic expressions such as x + 1/x, √x + 3, and ∛y + y². Do you know that you can write x + 1/x = x + x⁻¹? Here, the exponent of the second term, that is, x⁻¹ is -1, which is not a whole number. So, this algebraic expression is not a polynomial.

Again, √x + 3 can be written as x^(1/2) + 3. Here the exponent of x is 1/2, which is not a whole number. So, is √x + 3 a polynomial? No, it is not. What about ∛y + y²? It is also not a polynomial. Why? Because ∛y can be written as y^(1/3), and 1/3 is not a whole number. So, the key point is: a polynomial must have only whole number exponents (including zero) on the variable.

So, my dear students, let me make this very clear: a polynomial is an algebraic expression where all exponents of the variable are whole numbers (0, 1, 2, 3, and so on). If any exponent is negative or a fraction, then it is not a polynomial.

Now, if the variable in a polynomial is x, we may denote the polynomial by p(x), or q(x), or r(x), and so on. So, for example, we may write:

p(x) = 2x² + 5x - 3

q(x) = x³ - 1

r(y) = y³ + y + 1

s(u) = 2 - u - u² + 6u⁵

A polynomial can have any finite number of terms. For instance, x¹⁵⁰ + x¹⁴⁹ + ... + x² + x + 1 is a polynomial with 151 terms. The dots mean we continue in the same pattern.

Now, consider the polynomials 2x, 2, 5x³, -5x², y, and u⁴. Do you see that each of these polynomials has only one term? Polynomials having only one term are called monomials. The word 'mono' means 'one'. So, monomial means one term.

Now observe each of the following polynomials:

p(x) = x + 1, q(x) = x² - x, r(y) = y⁹ + 1, t(u) = u¹⁵ - u²

How many terms are there in each of these? Each of these polynomials has only two terms. Polynomials having only two terms are called binomials. The word 'bi' means 'two'.

Similarly, polynomials having only three terms are called trinomials. The word 'tri' means 'three'. Some examples of trinomials are:

p(x) = x + x² + π, q(x) = √2 + x - x², r(u) = u + u² - 2, t(y) = y⁴ + y + 5.

Now, let us learn about the degree of a polynomial. Look at the polynomial p(x) = 3x⁷ - 4x⁶ + x + 9. What is the term with the highest power of x? It is 3x⁷. The exponent of x in this term is 7. Similarly, in the polynomial q(y) = 5y⁶ - 4y² - 6, the term with the highest power of y is 5y⁶, and the exponent of y in this term is 6. We call the highest power of the variable in a polynomial as the degree of the polynomial. So, the degree of the polynomial 3x⁷ - 4x⁶ + x + 9 is 7, and the degree of the polynomial 5y⁶ - 4y² - 6 is 6. The degree of a non-zero constant polynomial is zero. This is because any non-zero constant can be written as a constant times x⁰, and the exponent is 0.

Now, let us look at Example 1 from your textbook to understand how to find the degree of a polynomial.

Example 1: Find the degree of each of the polynomials given below:

(i) x⁵ - x⁴ + 3

(ii) 2 - y² - y³ + 2y⁸

(iii) 2

Solution: (i) The highest power of the variable is 5. So, the degree of the polynomial is 5.

(ii) The highest power of the variable is 8. So, the degree of the polynomial is 8.

(iii) The only term here is 2, which can be written as 2x⁰. So the exponent of x is 0. Therefore, the degree of the polynomial is 0.

So, students, you must remember that to find the degree of a polynomial, we look for the term with the highest exponent of the variable. That exponent is the degree of the polynomial.

Now, let us talk about different types of polynomials based on their degree.

Now observe the polynomials p(x) = 4x + 5, q(y) = 2y, r(t) = t + √2, and s(u) = 3 - u. Do you see anything common among all of them? The degree of each of these polynomials is one. A polynomial of degree one is called a linear polynomial. Some more linear polynomials in one variable are 2x - 1, √2 y + 1, and 2 - u. Now, try and find a linear polynomial in x with 3 terms. You would not be able to find it because a linear polynomial in x can have at most two terms. So, any linear polynomial in x will be of the form ax + b, where a and b are constants and a ≠ 0. Why is a ≠ 0? Because if a = 0, then the polynomial would become just b, which is a constant polynomial, not linear. Similarly, ay + b is a linear polynomial in y.

Now consider the polynomials: 2x² + 5, 5x² + 3x + π, x², and x² + (2/5)x. Do you agree that they are all of degree two? A polynomial of degree two is called a quadratic polynomial. Some examples of a quadratic polynomial are 5 - y², 4y + 5y², and 6 - y - y². Can you write a quadratic polynomial in one variable with four different terms? You will find that a quadratic polynomial in one variable will have at most 3 terms. If you list a few more quadratic polynomials, you will find that any quadratic polynomial in x is of the form ax² + bx + c, where a ≠ 0, and a, b, c are constants. Similarly, a quadratic polynomial in y will be of the form ay² + by + c, provided a ≠ 0 and a, b, c are constants.

We call a polynomial of degree three a cubic polynomial. Some examples of a cubic polynomial in x are 4x³, 2x³ + 1, 5x³ + x², 6x³ - x, 6 - x³, and 2x³ + 4x² + 6x + 7. How many terms do you think a cubic polynomial in one variable can have? It can have at most 4 terms. These may be written in the form ax³ + bx² + cx + d, where a ≠ 0, and a, b, c, and d are constants.

Now, that you have seen what a polynomial of degree 1, degree 2, or degree 3 looks like, can you write down a polynomial in one variable of degree n for any natural number n? A polynomial in one variable x of degree n is an expression of the form:

aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀

where a₀, a₁, a₂, ..., aₙ are constants and aₙ ≠ 0.

In particular, if a₀ = a₁ = a₂ = a₃ = ... = aₙ = 0 (all the constants are zero), we get the zero polynomial, which is denoted by 0. What is the degree of the zero polynomial? The degree of the zero polynomial is not defined. This is an important point to remember.

So far we have dealt with polynomials in one variable only. We can also have polynomials in more than one variable. For example, x² + y² + xyz (where variables are x, y, and z) is a polynomial in three variables. Similarly, p² + q¹⁰ + r (where the variables are p, q, and r), and u³ + v² (where the variables are u and v) are polynomials in three and two variables, respectively. You will be studying such polynomials in detail later.

Now, let us summarize what we have learned so far in this section. We learned that a polynomial is an algebraic expression with whole number exponents. We learned about terms and coefficients of a polynomial. We learned about monomials (one term), binomials (two terms), and trinomials (three terms). We learned about the degree of a polynomial, and we learned about linear polynomials (degree 1), quadratic polynomials (degree 2), and cubic polynomials (degree 3). We also learned that the general form of a polynomial of degree n is aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀.

Now, let us move on to Section 2.3, which is about "Zeroes of a Polynomial".

Consider the polynomial p(x) = 5x³ - 2x² + 3x - 2.

If we replace x by 1 everywhere in p(x), we get:

p(1) = 5 × (1)³ - 2 × (1)² + 3 × (1) - 2 = 5 - 2 + 3 - 2 = 4

So, we say that the value of p(x) at x = 1 is 4.

Similarly, p(0) = 5(0)³ - 2(0)² + 3(0) - 2 = -2.

Can you find p(-1)? Let me tell you: p(-1) = 5(-1)³ - 2(-1)² + 3(-1) - 2 = 5(-1) - 2(1) - 3 - 2 = -5 - 2 - 3 - 2 = -12.

Now, let us look at Example 2 from your textbook.

Example 2: Find the value of each of the following polynomials at the indicated value of variables:

(i) p(x) = 5x² - 3x + 7 at x = 1.

(ii) q(y) = 3y³ - 4y + √11 at y = 2.

(iii) p(t) = 4t⁴ + 5t³ - t² + 6 at t = a.

Solution: (i) p(x) = 5x² - 3x + 7

The value of the polynomial p(x) at x = 1 is given by:

p(1) = 5(1)² - 3(1) + 7 = 5 - 3 + 7 = 9

(ii) q(y) = 3y³ - 4y + √11

The value of the polynomial q(y) at y = 2 is given by:

q(2) = 3(2)³ - 4(2) + √11 = 3 × 8 - 8 + √11 = 24 - 8 + √11 = 16 + √11

(iii) p(t) = 4t⁴ + 5t³ - t² + 6

The value of the polynomial p(t) at t = a is given by:

p(a) = 4a⁴ + 5a³ - a² + 6

Now, consider the polynomial p(x) = x - 1.

What is p(1)? Note that: p(1) = 1 - 1 = 0.

As p(1) = 0, we say that 1 is a zero of the polynomial p(x).

Similarly, you can check that 2 is a zero of q(x), where q(x) = x - 2.

In general, we say that a zero of a polynomial p(x) is a number c such that p(c) = 0.

You must have observed that the zero of the polynomial x - 1 is obtained by equating it to 0, that is, x - 1 = 0, which gives x = 1. We say p(x) = 0 is a polynomial equation, and 1 is the root of the polynomial equation p(x) = 0. So we say 1 is the zero of the polynomial x - 1, or a root of the polynomial equation x - 1 = 0.

Now, consider the constant polynomial 5. Can you tell what its zero is? It has no zero because replacing x by any number in 5x⁰ still gives us 5. In fact, a non-zero constant polynomial has no zero. What about the zeroes of the zero polynomial? By convention, every real number is a zero of the zero polynomial. This is an interesting fact!

Now, let us look at Example 3.

Example 3: Check whether -2 and 2 are zeroes of the polynomial x + 2.

Solution: Let p(x) = x + 2.

Then p(2) = 2 + 2 = 4, p(-2) = -2 + 2 = 0

Therefore, -2 is a zero of the polynomial x + 2, but 2 is not.

Example 4: Find a zero of the polynomial p(x) = 2x + 1.

Solution: Finding a zero of p(x) is the same as solving the equation p(x) = 0.

Now, 2x + 1 = 0 gives us x = -1/2.

So, -1/2 is a zero of the polynomial 2x + 1.

Now, if p(x) = ax + b, where a ≠ 0, is a linear polynomial, how can we find a zero of p(x)? Example 4 may have given you some idea. Finding a zero of the polynomial p(x) amounts to solving the polynomial equation p(x) = 0.

Now, p(x) = 0 means ax + b = 0, where a ≠ 0.

So, ax = -b

That is, x = -b/a.

So, x = -b/a is the only zero of p(x). In other words, a linear polynomial has one and only one zero.

Now we can say that 1 is the zero of x - 1, and -2 is the zero of x + 2.

Let us look at Example 5.

Example 5: Verify whether 2 and 0 are zeroes of the polynomial x² - 2x.

Solution: Let p(x) = x² - 2x

Then p(2) = 2² - 4 = 4 - 4 = 0

and p(0) = 0 - 0 = 0

Hence, 2 and 0 are both zeroes of the polynomial x² - 2x.

Now, let us list our observations:

(i) A zero of a polynomial need not be 0. For example, in the polynomial x - 1, the zero is 1, which is not 0.

(ii) 0 may be a zero of a polynomial. For example, in the polynomial x² - 2x, 0 is a zero.

(iii) Every linear polynomial has one and only one zero. This is because if p(x) = ax + b, then the only solution to p(x) = 0 is x = -b/a.

(iv) A polynomial can have more than one zero. For example, the polynomial x² - 2x has two zeroes: 0 and 2.

So, students, this is all about zeroes of a polynomial. Remember, finding the zero of a polynomial is the same as solving the polynomial equation p(x) = 0.

Now, let us move on to Section 2.4, which is about "Factorisation of Polynomials".

This is a very important section where we will learn about the Factor Theorem, which is extremely useful for factorising polynomials.

Let us first understand what it means for one polynomial to be a factor of another. If p(x) is a polynomial, and we can write p(x) = (x - a) × q(x) for some polynomial q(x), then (x - a) is a factor of p(x). In other words, if when we divide p(x) by (x - a), the remainder is 0, then (x - a) is a factor of p(x).

Now, let me state the Factor Theorem for you:

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) if p(a) = 0, and

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

This is a very powerful theorem. Let me explain the proof:

Proof: By the Remainder Theorem, which we will study in detail later, we have:

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

(i) If p(a) = 0, then p(x) = (x - a) q(x), which shows that (x - a) is a factor of p(x).

(ii) Since (x - a) is a factor of p(x), p(x) = (x - a) g(x) for some polynomial g(x). In this case, p(a) = (a - a) g(a) = 0.

So, the key idea is: to check if (x - a) is a factor of p(x), we just need to check if p(a) = 0. If p(a) = 0, then (x - a) is a factor; if p(a) ≠ 0, then (x - a) is not a factor.

Now, let us look at some examples to understand how to use the Factor Theorem.

Example 6: Examine whether x + 2 is a factor of x³ + 3x² + 5x + 6 and of 2x + 4.

Solution: The zero of x + 2 is -2. Let p(x) = x³ + 3x² + 5x + 6 and s(x) = 2x + 4.

Then, p(-2) = (-2)³ + 3(-2)² + 5(-2) + 6 = -8 + 12 - 10 + 6 = 0

So, by the Factor Theorem, x + 2 is a factor of x³ + 3x² + 5x + 6.

Again, s(-2) = 2(-2) + 4 = 0

So, x + 2 is a factor of 2x + 4. In fact, you can check this without applying the Factor Theorem, since 2x + 4 = 2(x + 2).

Example 7: Find the value of k, if x - 1 is a factor of 4x³ + 3x² - 4x + k.

Solution: As x - 1 is a factor of p(x) = 4x³ + 3x² - 4x + k, we have p(1) = 0.

Now, p(1) = 4(1)³ + 3(1)² - 4(1) + k = 4 + 3 - 4 + k = 3 + k

So, 3 + k = 0

Therefore, k = -3.

Now, we will use the Factor Theorem to factorise some polynomials of degree 2 and 3. You are already familiar with the factorisation of a quadratic polynomial like x² + lx + m. You had factorised it by splitting the middle term lx as ax + bx so that ab = m. Then x² + lx + m = (x + a)(x + b). We shall now try to factorise quadratic polynomials of the type ax² + bx + c, where a ≠ 0 and a, b, c are constants.

Factorisation of the polynomial ax² + bx + c by splitting the middle term is as follows:

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.

Example 8: Factorise 6x² + 17x + 5 by splitting the middle term, and by using the Factor Theorem.

Solution 1: (By splitting method): 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.

So, let us look for the pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5 and 6. Of these pairs, 2 and 15 will give us p + q = 17.

So, 6x² + 17x + 5 = 6x² + (2 + 15)x + 5 = 6x² + 2x + 15x + 5 = 2x(3x + 1) + 5(3x + 1) = (3x + 1)(2x + 5)

Solution 2: (Using the Factor Theorem)

6x² + 17x + 5 = 6(x² + 17/6 x + 5/6) = 6 p(x), say. If a and b are the zeroes of p(x), then 6x² + 17x + 5 = 6(x - a)(x - b). So, ab = 5/6. Let us look at some possibilities for a and b. They could be ±1/2, ±1/3, ±5/3, ±5/2, ±1. Now, p(1/2) = 1/4 + 17/6(1/2) + 5/6 ≠ 0. But p(-1/3) = 0. So, (x + 1/3) is a factor of p(x). Similarly, by trial, you can find that (x + 5/2) is a factor of p(x).

Therefore, 6x² + 17x + 5 = 6(x + 1/3)(x + 5/2) = 6(3x + 1/3)(2x + 5/2) = (3x + 1)(2x + 5)

For the example above, the use of the splitting method appears more efficient. However, let us consider another example.

Example 9: Factorise y² - 5y + 6 by using the Factor Theorem.

Solution: Let p(y) = y² - 5y + 6. Now, if p(y) = (y - a)(y - b), you know that the constant term will be ab. So, ab = 6. So, to look for the factors of p(y), we look at the factors of 6.

The factors of 6 are 1, 2, and 3.

Now, p(2) = 2² - (5 × 2) + 6 = 4 - 10 + 6 = 0

So, y - 2 is a factor of p(y).

Also, p(3) = 3² - (5 × 3) + 6 = 9 - 15 + 6 = 0

So, y - 3 is also a factor of y² - 5y + 6.

Therefore, y² - 5y + 6 = (y - 2)(y - 3)

Note that y² - 5y + 6 can also be factorised by splitting the middle term -5y.

Now, let us consider factorising cubic polynomials. Here, the splitting method will not be appropriate to start with. We need to find at least one factor first, as you will see in the following example.

Example 10: Factorise x³ - 23x² + 142x - 120.

Solution: Let p(x) = x³ - 23x² + 142x - 120

We shall now look for all the factors of -120. Some of these are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60.

By trial, we find that p(1) = 0. So x - 1 is a factor of p(x).

Now we see that:

x³ - 23x² + 142x - 120 = x³ - x² - 22x² + 22x + 120x - 120 = x²(x - 1) - 22x(x - 1) + 120(x - 1) = (x - 1)(x² - 22x + 120)

We could have also got this by dividing p(x) by x - 1.

Now x² - 22x + 120 can be factorised either by splitting the middle term or by using the Factor Theorem. By splitting the middle term, we have:

x² - 22x + 120 = x² - 12x - 10x + 120 = x(x - 12) - 10(x - 12) = (x - 12)(x - 10)

So, x³ - 23x² + 142x - 120 = (x - 1)(x - 10)(x - 12)

So, students, you can see how useful the Factor Theorem is for factorising polynomials, especially cubic polynomials. The key idea is to find one factor first by trial, and then use that to factorise the rest of the polynomial.

Now, let us move on to Section 2.5, which is about "Algebraic Identities".

From your earlier classes, you may recall that an algebraic identity is an algebraic equation that is true for all values of the variables occurring in it. You have studied the following algebraic identities in earlier classes:

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

You must have also used some of these algebraic identities to factorise the algebraic expressions. You can also see their utility in computations.

Now, let us look at some examples to see how these identities are used.

Example 11: Find the following products using appropriate identities:

(i) (x + 3)(x + 3)

(ii) (x - 3)(x + 5)

Solution: (i) Here we can use Identity I: (x + y)² = x² + 2xy + y². Putting y = 3 in it, we get:

(x + 3)(x + 3) = (x + 3)² = x² + 2(x)(3) + (3)² = x² + 6x + 9

(ii) Using Identity IV above, that is, (x + a)(x + b) = x² + (a + b)x + ab, we have:

(x - 3)(x + 5) = x² + (-3 + 5)x + (-3)(5) = x² + 2x - 15

Example 12: Evaluate 105 × 106 without multiplying directly.

Solution:

105 × 106 = (100 + 5) × (100 + 6) = (100)² + (5 + 6)(100) + (5 × 6), using Identity IV = 10000 + 1100 + 30 = 11130

This is a very useful technique. Instead of multiplying two numbers directly, we express them as sums or differences of numbers that are easy to square or multiply, and then use the algebraic identities.

Now, let us look at Example 13 to see how these identities are used in factorisation.

Example 13: Factorise:

(i) 49a² + 70ab + 25b²

(ii) (25/4)x² - (y²/9)

Solution: (i) Here you can see that:

49a² = (7a)², 25b² = (5b)², 70ab = 2(7a)(5b)

Comparing the given expression with x² + 2xy + y², we observe that x = 7a and y = 5b.

Using Identity I, we get:

49a² + 70ab + 25b² = (7a + 5b)² = (7a + 5b)(7a + 5b)

(ii) We have (25/4)x² - (y²/9) = ((5/2)x)² - ((y/3))²

Now comparing it with Identity III, we get:

(25/4)x² - (y²/9) = ((5/2)x)² - ((y/3))² = ((5/2)x + (y/3))((5/2)x - (y/3))

So far, all our identities involved products of binomials. Let us now extend Identity I to a trinomial x + y + z. We shall compute (x + y + z)² by using Identity I.

Let x + y = t. Then:

(x + y + z)² = (t + z)² = t² + 2tz + z² (Using Identity I) = (x + y)² + 2(x + y)z + z² (Substituting the value of t) = x² + 2xy + y² + 2xz + 2yz + z² (Using Identity I) = x² + y² + z² + 2xy + 2yz + 2zx (Rearranging the terms)

So, we get the following identity:

Identity V: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

We call the right-hand side expression the expanded form of the left-hand side expression. Note that the expansion of (x + y + z)² consists of three square terms and three product terms.

Example 14: Write (3a + 4b + 5c)² in expanded form.

Solution: Comparing the given expression with (x + y + z)², we find that:

x = 3a, y = 4b, and z = 5c.

Therefore, using Identity V, we have:

(3a + 4b + 5c)² = (3a)² + (4b)² + (5c)² + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a) = 9a² + 16b² + 25c² + 24ab + 40bc + 30ac

Example 15: Expand (4a - 2b - 3c)².

Solution: Using Identity V, we have:

(4a - 2b - 3c)² = [4a + (-2b) + (-3c)]² = (4a)² + (-2b)² + (-3c)² + 2(4a)(-2b) + 2(-2b)(-3c) + 2(-3c)(4a) = 16a² + 4b² + 9c² - 16ab + 12bc - 24ac

Example 16: Factorise 4x² + y² + z² - 4xy - 2yz + 4xz.

Solution: We have:

4x² + y² + z² - 4xy - 2yz + 4xz = (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)

So far, we have dealt with identities involving second-degree terms. Now let us extend Identity I to compute (x + y)³. We have:

(x + y)³ = (x + y)(x + y)² = (x + y)(x² + 2xy + y²) = x(x² + 2xy + y²) + y(x² + 2xy + y²) = x³ + 2x²y + xy² + x²y + 2xy² + y³ = x³ + 3x²y + 3xy² + y³ = x³ + y³ + 3xy(x + y)

So, we get the following identity:

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

Also, by replacing y by -y in Identity VI, we get:

Identity VII: (x - y)³ = x³ - y³ - 3xy(x - y) = x³ - 3x²y + 3xy² - y³

Example 17: Write the following cubes in the expanded form:

(i) (3a + 4b)³

(ii) (5p - 3q)³

Solution: (i) Comparing the given expression with (x + y)³, we find that:

x = 3a and y = 4b.

So, using Identity VI, we have:

(3a + 4b)³ = (3a)³ + (4b)³ + 3(3a)(4b)(3a + 4b) = 27a³ + 64b³ + 108a²b + 144ab²

(ii) Comparing the given expression with (x - y)³, we find that:

x = 5p, y = 3q.

So, using Identity VII, we have:

(5p - 3q)³ = (5p)³ - (3q)³ - 3(5p)(3q)(5p - 3q) = 125p³ - 27q³ - 225p²q + 135pq²

Example 18: Evaluate each of the following using suitable identities:

(i) (104)³

(ii) (999)³

Solution: (i) We have:

(104)³ = (100 + 4)³ = (100)³ + (4)³ + 3(100)(4)(100 + 4) (Using Identity VI) = 1000000 + 64 + 124800 = 1124864

(ii) We have:

(999)³ = (1000 - 1)³ = (1000)³ - (1)³ - 3(1000)(1)(1000 - 1) (Using Identity VII) = 1000000000 - 1 - 2997000 = 997002999

Now, let us look at Example 19 to see how we can factorise expressions that are sums of cubes.

Example 19: Factorise 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)

Now consider (x + y + z)(x² + y² + z² - xy - yz - zx). On expanding, we get the product as:

x(x² + y² + z² - xy - yz - zx) + y(x² + y² + z² - xy - yz - zx) + z(x² + y² + z² - xy - yz - zx) = x³ + xy² + xz² - x²y - xyz - zx² + x²y + y³ + yz² - xy² - y²z - xyz + x²z + y²z + z³ - xyz - yz² - xz² = x³ + y³ + z³ - 3xyz (On simplification)

So, we obtain the following identity:

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

Example 20: Factorise: 8x³ + y³ + 27z³ - 18xyz

Solution: Here, we have:

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)

Now, students, we have covered all the important concepts from this chapter. Let me now give you a comprehensive summary of everything we have learned in this chapter.

SUMMARY:

In this chapter, you have studied the following points:

1. A polynomial p(x) in one variable x is an algebraic expression in x of the form p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₂x² + a₁x + a₀, where a₀, a₁, a₂, ..., aₙ are constants and aₙ ≠ 0. a₀, a₁, a₂, ..., aₙ are respectively the coefficients of x⁰, x, x², ..., xⁿ, and n is called the degree of the polynomial. Each of aₙxⁿ, aₙ₋₁xⁿ⁻¹, ..., a₀, with aₙ ≠ 0, is called a term of the polynomial p(x).

2. A polynomial of one term is called a monomial.

3. A polynomial of two terms is called a binomial.

4. A polynomial of three terms is called a trinomial.

5. A polynomial of degree one is called a linear polynomial.

6. A polynomial of degree two is called a quadratic polynomial.

7. A polynomial of degree three is called a cubic polynomial.

8. A real number 'a' is a zero of a polynomial p(x) if p(a) = 0. In this case, a is also called a root of the equation p(x) = 0.

9. Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial has no zero, and every real number is a zero of the zero polynomial.

10. Factor Theorem: (x - a) is a factor of the polynomial p(x) if p(a) = 0. Also, if (x - a) is a factor of p(x), then p(a) = 0.

11. (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

12. (x + y)³ = x³ + y³ + 3xy(x + y)

13. (x - y)³ = x³ - y³ - 3xy(x - y)

14. x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

Now, my dear students, I hope you have understood all the concepts in this chapter clearly. Remember, mathematics is not about memorising formulas but about understanding the concepts and applying them to solve problems. Practice as many problems as you can, and you will surely master this chapter.

Thank you for your attention, and I will see you in the next lesson. Keep studying hard!