Polynomials

Welcome dear students! Today we are going to learn about Polynomials from Class 9 Maths. You have studied algebraic expressions, their addition, subtraction, multiplication and division in earlier classes. You also have studied how to factorise some algebraic expressions. You may recall the algebraic identities: (x + y)² = x² + 2xy + y², (x – y)² = x² – 2xy + y², and x² – y² = (x + y)(x – y) and their use in factorisation. In this chapter, we shall start our study with a particular type of algebraic expression, called polynomial, and the terminology related to it. We shall also study the Remainder Theorem and Factor Theorem and their use in the factorisation of polynomials. In addition to the above, we shall study some more algebraic identities and their use in factorisation and in evaluating some given expressions. [CHECKPOINT] Let us begin by recalling that a variable is denoted by a symbol that can take any real value. We use the letters x, y, z, et cetera, to denote variables. Notice that 2x, 3x, –x, –1/2 x are algebraic expressions. All these expressions are of the form a constant multiplied by x. Now suppose we want to write an expression which is a constant multiplied by a variable and we do not know what the constant is. In such cases, we write the constant as a, b, c, et cetera. So the expression will be ax, say. However, there is a 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. [CHECKPOINT] Now, consider a square of side 3 units. In Figure 2.1, we see a square with each side measuring 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 multiplied by 3, that is, 12 units. What will be the perimeter if each side of the square is 10 units? The perimeter is 4 multiplied by 10, that is, 40 units. In case the length of each side is x units, as shown in Figure 2.2 where a square PQRS has side length x, the perimeter is given by 4x units. So, as the length of the side varies, the perimeter varies. Can you find the area of the square PQRS? It is x multiplied by x, which equals x² square units. x² is an algebraic expression. You are also familiar with other algebraic expressions like 2x, x² + 2x, x³ – x² + 4x + 7. Note that, 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. [CHECKPOINT] 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 equals 1. Do you know the coefficient of x in x² – x + 7? It is –1. 2 is also a polynomial. In fact, 2, –5, 7, et cetera, 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. [CHECKPOINT] Now, consider algebraic expressions such as x + 1/x, 3√x, and y + 2/y. Do you know that you can write x + 1/x as 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, 3√x can be written as 3x^(1/2). Here the exponent of x is 1/2, which is not a whole number. So, is 3√x a polynomial? No, it is not. What about 3/y + y²? It is also not a polynomial. Why? Because the exponent of y in the first term is –1, which is not a whole number. If the variable in a polynomial is x, we may denote the polynomial by p(x), or q(x), or r(x), et cetera. 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 + dot dot dot + x² + x + 1 is a polynomial with 151 terms. [CHECKPOINT] 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. Mono means one. 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. Bi means two. Similarly, polynomials having only three terms are called trinomials. 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, 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. [CHECKPOINT] 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. 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, 2y + 1, 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 is not equal to 0. Why? Because if a is 0, the degree becomes 0, making it a constant polynomial. Similarly, ay + b is a linear polynomial in y. [CHECKPOINT] 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 is not equal to 0 and a, b, c are constants. Similarly, quadratic polynomial in y will be of the form ay² + by + c, provided a is not equal to 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, 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 is not equal to 0 and a, b, c and d are constants. [CHECKPOINT] 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 + dot dot dot + a₁x + a₀ where a₀, a₁, a₂, dot dot dot, aₙ are constants and aₙ is not equal to 0. In particular, if a₀ = a₁ = a₂ = a₃ = dot dot dot = aₙ = 0, meaning 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. 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. [CHECKPOINT] Let us now move to Exercise 2.1. Question 1: Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (i) 4x² – 3x + 7. This is a polynomial in one variable x because all exponents of x are whole numbers. (ii) y² + 2. This is a polynomial in one variable y because the exponent of y is a whole number. (iii) 3√t + t√2. This is not a polynomial because 3√t can be written as 3t^(1/2), and the exponent 1/2 is not a whole number. (iv) y + 2/y. This is not a polynomial because 2/y is 2y⁻¹, and the exponent –1 is not a whole number. (v) x¹⁰ + y³ + t⁵⁰. This is not a polynomial in one variable because it contains three variables x, y, and t. Question 2: Write the coefficients of x² in each of the following: (i) 2 + x² + x. The coefficient of x² is 1. (ii) 2 – x² + x³. The coefficient of x² is –1. (iii) (π/2)x² + x. The coefficient of x² is π/2. (iv) √2 x – 1. The coefficient of x² is 0 because there is no x² term. [CHECKPOINT] Question 3: Give one example each of a binomial of degree 35, and of a monomial of degree 100. A binomial of degree 35 is x³⁵ + 1. A monomial of degree 100 is 5x¹⁰⁰. Question 4: Write the degree of each of the following polynomials: (i) 5x³ + 4x² + 7x. The highest power is 3, so degree is 3. (ii) 4 – y². The highest power is 2, so degree is 2. (iii) 5t – 7. The highest power is 1, so degree is 1. (iv) 3. This is a constant, so it can be written as 3x⁰. The degree is 0. Question 5: Classify the following as linear, quadratic and cubic polynomials: (i) x² + x. Degree is 2, so it is quadratic. (ii) x – x³. Degree is 3, so it is cubic. (iii) y + y² + 4. Degree is 2, so it is quadratic. (iv) 1 + x. Degree is 1, so it is linear. (v) 3t. Degree is 1, so it is linear. (vi) r². Degree is 2, so it is quadratic. (vii) 7x³. Degree is 3, so it is cubic. [CHECKPOINT] Now, let us proceed to Section 2.3, 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 multiplied by (1)³ – 2 multiplied by (1)² + 3 multiplied by (1) – 2, which equals 5 – 2 + 3 – 2, which is 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 us work through 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. Solution: 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 at y = 2. Solution: The value of the polynomial q(y) at y = 2 is given by q(2) = 3(2)³ – 4(2) + 11 = 24 – 8 + 11 = 27. (iii) p(t) = 4t⁴ + 5t³ – t² + 6 at t = a. Solution: The value of the polynomial p(t) at t = a is given by p(a) = 4a⁴ + 5a³ – a² + 6. [CHECKPOINT] 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. [CHECKPOINT] 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, and 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, a is not equal to 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, a is not equal to 0. So, ax = –b, that is, x = –b/a. So, x = –b/a is the only zero of p(x), that is, 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. [CHECKPOINT] 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. Let us now list our observations: First, a zero of a polynomial need not be 0. Second, 0 may be a zero of a polynomial. Third, every linear polynomial has one and only one zero. Fourth, a polynomial can have more than one zero. Now let us move to Exercise 2.2. Question 1: Find the value of the polynomial 5x – 4x² + 3 at (i) x = 0, (ii) x = –1, (iii) x = 2. Solution: (i) At x = 0, the value is 5(0) – 4(0)² + 3 = 3. (ii) At x = –1, the value is 5(–1) – 4(–1)² + 3 = –5 – 4 + 3 = –6. (iii) At x = 2, the value is 5(2) – 4(2)² + 3 = 10 – 16 + 3 = –3. [CHECKPOINT] Question 2: Find p(0), p(1) and p(2) for each of the following polynomials: (i) p(y) = y² – y + 1. p(0) = 0 – 0 + 1 = 1. p(1) = 1 – 1 + 1 = 1. p(2) = 4 – 2 + 1 = 3. (ii) p(t) = 2 + t + 2t² – t³. p(0) = 2. p(1) = 2 + 1 + 2 – 1 = 4. p(2) = 2 + 2 + 8 – 8 = 4. (iii) p(x) = x³. p(0) = 0. p(1) = 1. p(2) = 8. (iv) p(x) = (x – 1)(x + 1). p(0) = (–1)(1) = –1. p(1) = (0)(2) = 0. p(2) = (1)(3) = 3. Question 3: Verify whether the following are zeroes of the polynomial, indicated against them. (i) p(x) = 3x + 1, x = –1/3. p(–1/3) = 3(–1/3) + 1 = –1 + 1 = 0. Yes, it is a zero. (ii) p(x) = 5x – π, x = 4/5. p(4/5) = 5(4/5) – π = 4 – π, which is not 0. So, no. (iii) p(x) = x² – 1, x = 1, –1. p(1) = 1 – 1 = 0. p(–1) = 1 – 1 = 0. Both are zeroes. (iv) p(x) = (x + 1)(x – 2), x = –1, 2. p(–1) = 0. p(2) = 0. Both are zeroes. (v) p(x) = x², x = 0. p(0) = 0. Yes. (vi) p(x) = lx + m, x = –m/l. p(–m/l) = l(–m/l) + m = –m + m = 0. Yes. (vii) p(x) = 3x² – 1, x = 1/√3, –1/√3. p(1/√3) = 3(1/3) – 1 = 0. p(–1/√3) = 3(1/3) – 1 = 0. Both are zeroes. (viii) p(x) = 2x + 1, x = 1/2. p(1/2) = 2(1/2) + 1 = 2, not 0. So, no. [CHECKPOINT] Question 4: Find the zero of the polynomial in each of the following cases: (i) p(x) = x + 5. Set x + 5 = 0, so x = –5. (ii) p(x) = x – 5. Set x – 5 = 0, so x = 5. (iii) p(x) = 2x + 5. Set 2x + 5 = 0, so 2x = –5, x = –5/2. (iv) p(x) = 3x – 2. Set 3x – 2 = 0, so 3x = 2, x = 2/3. (v) p(x) = 3x. Set 3x = 0, so x = 0. (vi) p(x) = ax, a is not equal to 0. Set ax = 0, so x = 0. (vii) p(x) = cx + d, c is not equal to 0. Set cx + d = 0, so cx = –d, x = –d/c. Now, let us proceed to Section 2.4, Factorisation of Polynomials. Let us now look at the situation of Example 5 above more closely. It tells us that since the remainder is 0, (2t + 1) is a factor of q(t), that is, q(t) = (2t + 1)g(t) for some polynomial g(t). This is a particular case of the following theorem. 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). [CHECKPOINT] Proof: By the Remainder Theorem, 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 same polynomial g(x). In this case, p(a) = (a – a)g(a) = 0. 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, p(1) = 0. Now, p(1) = 4(1)³ + 3(1)² – 4(1) + k. So, 4 + 3 – 4 + k = 0, that is, k = –3. [CHECKPOINT] We will now 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 is not equal to 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. This will be clear from Example 8. 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 multiplied by 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). [CHECKPOINT] 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 plus or minus 1, plus or minus 1/2, plus or minus 1/3, plus or minus 5/3, plus or minus 5/2, plus or minus 5. Now, p(–1/2) = (–1/2)² + 17/6(–1/2) + 5/6 = 1/4 – 17/12 + 5/6 = (3 – 17 + 10)/12 = –4/12, which is not 0. But p(–1/3) = (–1/3)² + 17/6(–1/3) + 5/6 = 1/9 – 17/18 + 5/6 = (2 – 17 + 15)/18 = 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 multiplied by 2) + 6 = 4 – 10 + 6 = 0. So, y – 2 is a factor of p(y). Also, p(3) = 3² – (5 multiplied by 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. [CHECKPOINT] 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 plus or minus 1, plus or minus 2, plus or minus 3, plus or minus 4, plus or minus 5, plus or minus 6, plus or minus 8, plus or minus 10, plus or minus 12, plus or minus 15, plus or minus 20, plus or minus 24, plus or minus 30, plus or minus 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) by taking (x – 1) common. 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). [CHECKPOINT] Let us now move to Exercise 2.3. Question 1: Determine which of the following polynomials has (x + 1) a factor: (i) x³ + x² + x + 1. Let p(x) = x³ + x² + x + 1. Zero of x + 1 is –1. p(–1) = –1 + 1 – 1 + 1 = 0. Yes, it is a factor. (ii) x⁴ + x³ + x² + x + 1. p(–1) = 1 – 1 + 1 – 1 + 1 = 1. Not 0. No. (iii) x⁴ + 3x³ + 3x² + x + 1. p(–1) = 1 – 3 + 3 – 1 + 1 = 1. Not 0. No. (iv) x³ – x² – (√2 + 2)x + √2. p(–1) = –1 – 1 – (√2 + 2)(–1) + √2 = –2 + √2 + 2 + √2 = 2√2. Not 0. No. Question 2: Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1. Zero is –1. p(–1) = 2(–1) + 1 + 2 – 1 = –2 + 1 + 2 – 1 = 0. Yes. (ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2. Zero is –2. p(–2) = –8 + 12 – 6 + 1 = –1. Not 0. No. (iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3. Zero is 3. p(3) = 27 – 36 + 3 + 6 = 0. Yes. [CHECKPOINT] Question 3: Find the value of k, if x – 1 is a factor of p(x) in each of the following cases: (i) p(x) = x² + x + k. p(1) = 1 + 1 + k = 0, so k = –2. (ii) p(x) = 2x² + kx + 2. p(1) = 2 + k + 2 = 0, so k = –4. (iii) p(x) = kx² – √2 x + 1. p(1) = k – √2 + 1 = 0, so k = √2 – 1. (iv) p(x) = kx² – 3x + k. p(1) = k – 3 + k = 0, so 2k = 3, k = 3/2. Question 4: Factorise: (i) 12x² – 7x + 1. We need two numbers that add to –7 and multiply to 12. They are –3 and –4. So, 12x² – 3x – 4x + 1 = 3x(4x – 1) – 1(4x – 1) = (4x – 1)(3x – 1). (ii) 2x² + 7x + 3. Numbers are 6 and 1. So, 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1). (iii) 6x² + 5x – 6. Numbers are 9 and –4. So, 6x² + 9x – 4x – 6 = 3x(2x + 3) – 2(2x + 3) = (2x + 3)(3x – 2). (iv) 3x² – x – 4. Numbers are –4 and 3. So, 3x² – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1). [CHECKPOINT] Question 5: Factorise: (i) x³ – 2x² – x + 2. Let p(x) = x³ – 2x² – x + 2. Try x = 1: p(1) = 1 – 2 – 1 + 2 = 0. So (x – 1) is a factor. Divide or group: x²(x – 1) – x(x – 1) – 2(x – 1) = (x – 1)(x² – x – 2). Now factor x² – x – 2: (x – 2)(x + 1). So, (x – 1)(x – 2)(x + 1). (ii) x³ – 3x² – 9x – 5. Try x = –1: p(–1) = –1 – 3 + 9 – 5 = 0. So (x + 1) is a factor. Group: x²(x + 1) – 4x(x + 1) – 5(x + 1) = (x + 1)(x² – 4x – 5). Factor quadratic: (x – 5)(x + 1). So, (x + 1)²(x – 5). (iii) x³ + 13x² + 32x + 20. Try x = –1: p(–1) = –1 + 13 – 32 + 20 = 0. So (x + 1) is a factor. Group: x²(x + 1) + 12x(x + 1) + 20(x + 1) = (x + 1)(x² + 12x + 20). Factor quadratic: (x + 2)(x + 10). So, (x + 1)(x + 2)(x + 10). (iv) 2y³ + y² – 2y – 1. Try y = 1: p(1) = 2 + 1 – 2 – 1 = 0. So (y – 1) is a factor. Group: y²(2y + 1) – 1(2y + 1) = (2y + 1)(y² – 1) = (2y + 1)(y – 1)(y + 1). [CHECKPOINT] Now, let us proceed to Section 2.5, 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. Example 11: Find the following products using appropriate identities: (i) (x + 3)(x + 3). Solution: 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) (x – 3)(x + 5). Solution: 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. [CHECKPOINT] Example 12: Evaluate 105 multiplied by 106 without multiplying directly. Solution: 105 multiplied by 106 = (100 + 5) multiplied by (100 + 6) = (100)² + (5 + 6)(100) + (5 multiplied by 6), using Identity IV = 10000 + 1100 + 30 = 11130. You have seen some uses of the identities listed above in finding the product of some given expressions. These identities are useful in factorisation of algebraic expressions also, as you can see in the following examples. Example 13: Factorise: (i) 49a² + 70ab + 25b². Solution: 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) 25y²/4 – 4x²/9. Solution: We have 25y²/4 – 4x²/9 = (5y/2)² – (2x/3)². Now comparing it with Identity III, we get 25y²/4 – 4x²/9 = (5y/2 – 2x/3)(5y/2 + 2x/3). [CHECKPOINT] So far, all our identities involved products of binomials. Let us now extend the 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. Substituting the value of t, we get (x + y)² + 2(x + y)z + z². Using Identity I again, this becomes x² + 2xy + y² + 2xz + 2yz + z². Rearranging the terms, we get x² + y² + z² + 2xy + 2yz + 2zx. So, we get the following identity: Identity V: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx. Remark: 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. [CHECKPOINT] 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 the Identity VI, we get Identity VII: (x – y)³ = x³ – y³ – 3xy(x – y) = x³ – 3x²y + 3xy² – y³. [CHECKPOINT] Example 17: Write the following cubes in the expanded form: (i) (3a + 4b)³. Solution: 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) (5p – 3q)³. Solution: 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. Solution: We have (104)³ = (100 + 4)³ = (100)³ + (4)³ + 3(100)(4)(100 + 4), using Identity VI = 1000000 + 64 + 124800 = 1124864. (ii) (999)³. Solution: We have (999)³ = (1000 – 1)³ = (1000)³ – (1)³ – 3(1000)(1)(1000 – 1), using Identity VII = 1000000000 – 1 – 2997000 = 997002999. [CHECKPOINT] 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. On simplification, this equals x³ + y³ + z³ – 3xyz. So, we obtain the following identity: Identity VIII: x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx). [CHECKPOINT] 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 let us move to Exercise 2.4. Question 1: Use suitable identities to find the following products: (i) (x + 4)(x + 10) = x² + (4 + 10)x + 40 = x² + 14x + 40. (ii) (x + 8)(x – 10) = x² + (8 – 10)x – 80 = x² – 2x – 80. (iii) (3x + 4)(3x – 5) = (3x)² + (4 – 5)(3x) – 20 = 9x² – 3x – 20. (iv) (y² + 3/2)(y² – 3/2) = (y²)² – (3/2)² = y⁴ – 9/4. (v) (3 – 2x)(3 + 2x) = 3² – (2x)² = 9 – 4x². [CHECKPOINT] Question 2: Evaluate the following products without multiplying directly: (i) 103 multiplied by 107 = (100 + 3)(100 + 7) = 100² + (3 + 7)100 + 21 = 10000 + 1000 + 21 = 11021. (ii) 95 multiplied by 96 = (100 – 5)(100 – 4) = 100² – (5 + 4)100 + 20 = 10000 – 900 + 20 = 9120. (iii) 104 multiplied by 96 = (100 + 4)(100 – 4) = 100² – 4² = 10000 – 16 = 9984. Question 3: Factorise the following using appropriate identities: (i) 9x² + 6xy + y² = (3x)² + 2(3x)(y) + y² = (3x + y)². (ii) 4y² – 4y + 1 = (2y)² – 2(2y)(1) + 1² = (2y – 1)². (iii) x² – y²/100 = x² – (y/10)² = (x – y/10)(x + y/10). Question 4: Expand each of the following, using suitable identities: (i) (x + 2y + 4z)² = x² + 4y² + 16z² + 4xy + 16yz + 8zx. (ii) (2x – y + z)² = 4x² + y² + z² – 4xy – 2yz + 4zx. (iii) (–2x + 3y + 2z)² = 4x² + 9y² + 4z² – 12xy + 12yz – 8zx. (iv) (3a – 7b – c)² = 9a² + 49b² + c² – 42ab + 14bc – 6ac. (v) (–2x + 5y – 3z)² = 4x² + 25y² + 9z² – 20xy – 30yz + 12zx. (vi) (1/4 a – 1/2 b + 1)² = 1/16 a² + 1/4 b² + 1 – 1/4 ab – b + 1/2 a. [CHECKPOINT] Question 5: Factorise: (i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz = (2x)² + (3y)² + (–4z)² + 2(2x)(3y) + 2(3y)(–4z) + 2(–4z)(2x) = (2x + 3y – 4z)². (ii) 2x² + y² + 8z² – 2√2 xy + 4√2 yz – 8xz = (√2x)² + y² + (2√2z)² + 2(√2x)(–y) + 2(–y)(–2√2z) + 2(2√2z)(√2x) = (√2x – y + 2√2z)². Question 6: Write the following cubes in expanded form: (i) (2x + 1)³ = 8x³ + 1 + 3(2x)(1)(2x + 1) = 8x³ + 6x² + 6x + 1. (ii) (2a – 3b)³ = 8a³ – 27b³ – 3(2a)(3b)(2a – 3b) = 8a³ – 27b³ – 36a²b + 54ab². (iii) (3/2 x + 1)³ = 27/8 x³ + 1 + 3(3/2 x)(1)(3/2 x + 1) = 27/8 x³ + 27/4 x² + 9/2 x + 1. (iv) (x – 2/3 y)³ = x³ – 8/27 y³ – 3(x)(2/3 y)(x – 2/3 y) = x³ – 8/27 y³ – 2x²y + 4/3 xy². [CHECKPOINT] Question 7: Evaluate the following using suitable identities: (i) (99)³ = (100 – 1)³ = 1000000 – 1 – 300(99) = 1000000 – 1 – 29700 = 970299. (ii) (102)³ = (100 + 2)³ = 1000000 + 8 + 600(102) = 1000000 + 8 + 61200 = 1061208. (iii) (998)³ = (1000 – 2)³ = 1000000000 – 8 – 6000(998) = 1000000000 – 8 – 5988000 = 994011992. Question 8: Factorise each of the following: (i) 8a³ + b³ + 12a²b + 6ab² = (2a)³ + b³ + 3(2a)²(b) + 3(2a)(b)² = (2a + b)³. (ii) 8a³ – b³ – 12a²b + 6ab² = (2a)³ – b³ – 3(2a)²(b) + 3(2a)(b)² = (2a – b)³. (iii) 27 – 125a³ – 135a + 225a² = 3³ – (5a)³ – 3(3)²(5a) + 3(3)(5a)² = (3 – 5a)³. (iv) 64a³ – 27b³ – 144a²b + 108ab² = (4a)³ – (3b)³ – 3(4a)²(3b) + 3(4a)(3b)² = (4a – 3b)³. (v) 27p³ – 1/216 – 9/2 p² + 1/4 p = (3p)³ – (1/6)³ – 3(3p)²(1/6) + 3(3p)(1/6)² = (3p – 1/6)³. [CHECKPOINT] Question 9: Verify: (i) x³ + y³ = (x + y)(x² – xy + y²). Expanding right side: x(x² – xy + y²) + y(x² – xy + y²) = x³ – x²y + xy² + x²y – xy² + y³ = x³ + y³. Verified. (ii) x³ – y³ = (x – y)(x² + xy + y²). Expanding right side: x(x² + xy + y²) – y(x² + xy + y²) = x³ + x²y + xy² – x²y – xy² – y³ = x³ – y³. Verified. Question 10: Factorise each of the following: (i) 27y³ + 125z³ = (3y)³ + (5z)³ = (3y + 5z)(9y² – 15yz + 25z²). (ii) 64m³ – 343n³ = (4m)³ – (7n)³ = (4m – 7n)(16m² + 28mn + 49n²). Question 11: Factorise: 27x³ + y³ + z³ – 9xyz = (3x)³ + y³ + z³ – 3(3x)(y)(z) = (3x + y + z)(9x² + y² + z² – 3xy – yz – 3zx). [CHECKPOINT] Question 12: Verify that x³ + y³ + z³ – 3xyz = 1/2 (x + y + z)[(x – y)² + (y – z)² + (z – x)²]. Right side: 1/2 (x + y + z)[x² – 2xy + y² + y² – 2yz + z² + z² – 2zx + x²] = 1/2 (x + y + z)[2x² + 2y² + 2z² – 2xy – 2yz – 2zx] = (x + y + z)(x² + y² + z² – xy – yz – zx) = x³ + y³ + z³ – 3xyz. Verified. Question 13: If x + y + z = 0, show that x³ + y³ + z³ = 3xyz. Using Identity VIII, x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx). Since x + y + z = 0, the right side is 0. So x³ + y³ + z³ – 3xyz = 0, which means x³ + y³ + z³ = 3xyz. Question 14: Without actually calculating the cubes, find the value of each of the following: (i) (–12)³ + (7)³ + (5)³. Note that –12 + 7 + 5 = 0. So sum of cubes is 3(–12)(7)(5) = –1260. (ii) (28)³ + (–15)³ + (–13)³. Note that 28 – 15 – 13 = 0. So sum of cubes is 3(28)(–15)(–13) = 16380. [CHECKPOINT] Question 15: Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (i) Area: 25a² – 35a + 12. Factorise: 25a² – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3). So length and breadth could be 5a – 4 and 5a – 3. (ii) Area: 35y² + 13y – 12. Factorise: 35y² + 28y – 15y – 12 = 7y(5y + 4) – 3(5y + 4) = (5y + 4)(7y – 3). So length and breadth could be 5y + 4 and 7y – 3. Question 16: What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume: 3x² – 12x. Factorise: 3x(x – 4). So dimensions could be 3, x, and x – 4. (ii) Volume: 12ky² + 8ky – 20k. Factorise: 4k(3y² + 2y – 5) = 4k(3y² + 5y – 3y – 5) = 4k[y(3y + 5) – 1(3y + 5)] = 4k(3y + 5)(y – 1). So dimensions could be 4k, 3y + 5, and y – 1. [CHECKPOINT] Let us now review the summary of this chapter. In this chapter, you have studied the following points. First, a polynomial p(x) in one variable x is an algebraic expression in x of the form p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + dot dot dot + a₂x² + a₁x + a₀, where a₀, a₁, a₂, dot dot dot, aₙ are constants and aₙ is not equal to 0. a₀, a₁, a₂, dot dot dot, aₙ are respectively the coefficients of x⁰, x, x², dot dot dot, xⁿ, and n is called the degree of the polynomial. Each of aₙxⁿ, aₙ₋₁xⁿ⁻¹, dot dot dot, a₀, with aₙ is not equal to 0, is called a term of the polynomial p(x). Second, a polynomial of one term is called a monomial. Third, a polynomial of two terms is called a binomial. Fourth, a polynomial of three terms is called a trinomial. Fifth, a polynomial of degree one is called a linear polynomial. Sixth, a polynomial of degree two is called a quadratic polynomial. Seventh, a polynomial of degree three is called a cubic polynomial. Eighth, 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. Ninth, 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. [CHECKPOINT] Tenth, 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. Eleventh, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx. Twelfth, (x + y)³ = x³ + y³ + 3xy(x + y). Thirteenth, (x – y)³ = x³ – y³ – 3xy(x – y). Fourteenth, x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx). Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]