Polynomials

Hello students, welcome to today's mathematics lesson. I'm so happy to be here with you to learn about polynomials, which is Chapter 2 of your NCERT textbook. This is a very important chapter that builds upon what you studied in Class IX, and I assure you that by the end of this lesson, you will have a complete understanding of every concept related to polynomials. So let's begin our journey together.

So students, let's start by understanding what polynomials are. You have already studied polynomials in one variable and their degrees in Class IX. Let me remind you of the basics. If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial p(x). This is very important, so remember this definition well. The degree tells us what type of polynomial we are dealing with.

Now let me give you some examples so that this becomes crystal clear. The expression 4x + 2 is a polynomial in the variable x of degree 1 because the highest power of x is 1. Similarly, 2y² – 3y + 4 is a polynomial in the variable y of degree 2. The expression 5x³ – 4x² + x – √2 is a polynomial in the variable x of degree 3. And 7u⁶ – 3/2 u⁴ + 4u² + u – 8 is a polynomial in the variable u of degree 6. Now students, I want you to notice something important here. Expressions like 1/(x-1), √x + 2, and 1/(x² + 2x + 3) are NOT polynomials. Do you know why? Because in a polynomial, the exponents of the variable must be whole numbers (non-negative integers). In 1/(x-1), we have x in the denominator, which means we have x to the power of -1, which is not a whole number. Similarly, in √x, we have x to the power of 1/2, which is also not a whole number. So students, always remember that for an expression to be a polynomial, all exponents of the variable must be non-negative integers (0, 1, 2, 3, and so on).

Now let's talk about the different types of polynomials based on their degrees. A polynomial of degree 1 is called a linear polynomial. The word "linear" comes from "line" because the graph of a linear polynomial is a straight line, as you will see later. Some examples of linear polynomials are: 2x – 3, √3x + 5, y + √2, x – 2/11, 3z + 4, and 2/3 u + 1. Notice that in all these examples, the highest power of the variable is 1. On the other hand, polynomials like 2x + 5 – x² and x³ + 1 are NOT linear polynomials because in the first case, we have x² (degree 2), and in the second case, we have x³ (degree 3).

Now students, a polynomial of degree 2 is called a quadratic polynomial. The name "quadratic" has been derived from the word "quadrate," which means "square." This makes sense because the highest power is 2, which is related to finding the area of a square. Some examples of quadratic polynomials are: 2x² + 3x – 2/5, y² – 2, 2 – x² + √3x, u/3 – 2u² + 5, √5v² – 2/3 v, and 4z² + 1/7. In all these examples, the highest power of the variable is 2, and the coefficients are real numbers. More generally, students, any quadratic polynomial in x is of the form ax² + bx + c, where a, b, c are real numbers and a is not equal to 0. The condition a ≠ 0 is very important because if a were 0, then we would not have an x² term, and it would not be a quadratic polynomial anymore.

Now, a polynomial of degree 3 is called a cubic polynomial. The word "cubic" is related to "cube" because degree 3 is associated with the volume of a cube. Some examples of cubic polynomials are: 2 – x³, x³, √2 x³, 3 – x² + x³, and 3x³ – 2x² + x – 1. The most general form of a cubic polynomial is ax³ + bx² + cx + d, where a, b, c, d are real numbers and a ≠ 0.

Now students, let's understand what it means to find the value of a polynomial at a particular point. Consider the polynomial p(x) = x² – 3x – 4. If I ask you what is the value of this polynomial at x = 2, you simply substitute 2 in place of x everywhere in the expression. So p(2) = 2² – 3 × 2 – 4 = 4 – 6 – 4 = –6. This value –6 is called the value of the polynomial at x = 2. Similarly, p(0) is the value at x = 0, which is 0² – 3 × 0 – 4 = –4. In general, if p(x) is a polynomial in x and k is any real number, then the value obtained by replacing x by k in p(x) is called the value of p(x) at x = k, and it is denoted by p(k).

Now here's a very important concept: zeroes of a polynomial. Let me explain this with an example. What is the value of p(x) = x² – 3x – 4 at x = –1? We have p(–1) = (–1)² – 3 × (–1) – 4 = 1 + 3 – 4 = 0. Also, p(4) = 4² – 3 × 4 – 4 = 16 – 12 – 4 = 0. So when we substitute x = –1 and x = 4 in the polynomial, we get 0. These special values are called zeroes of the polynomial. More precisely, a real number k is said to be a zero of a polynomial p(x) if p(k) = 0. So students, –1 and 4 are the zeroes of the quadratic polynomial x² – 3x – 4.

Now you might wonder, why are zeroes so important? Let me tell you that zeroes of a polynomial are very special numbers. For a linear polynomial, we can easily find the zero. For example, if k is a zero of p(x) = 2x + 3, then p(k) = 0 gives us 2k + 3 = 0, which means k = –3/2. In general, if k is a zero of p(x) = ax + b, then p(k) = ak + b = 0, so k = –b/a. This means the zero of a linear polynomial is –(constant term)/(coefficient of x). Students, notice that the zero of a linear polynomial is directly related to its coefficients. Now the question is: does a similar relationship exist for quadratic polynomials and cubic polynomials? This is exactly what we will explore in this chapter. We will also study the division algorithm for polynomials, which is another important topic.

Now let's move to Section 2.2, which talks about the geometrical meaning of zeroes of a polynomial. This is where things become really interesting because we connect algebra with geometry.

You already know that a real number k is a zero of the polynomial p(x) if p(k) = 0. But why are zeroes so important from a geometrical perspective? Let me explain.

First, let's understand the geometrical representation of a linear polynomial. Consider the linear polynomial ax + b, where a is not equal to 0. You have studied in Class IX that the graph of y = ax + b is a straight line. For example, let's take y = 2x + 3. We can make a table of values: when x = –2, y = 2 × (–2) + 3 = –4 + 3 = –1. When x = 2, y = 2 × 2 + 3 = 4 + 3 = 7. So we have the points (–2, –1) and (2, 7) on this line. Now, if we draw this line on a graph paper, we will see that it intersects the x-axis at one point. Can you find where? The x-axis is where y = 0. So we need to find where 2x + 3 = 0, which gives x = –3/2. So the line intersects the x-axis at the point (–3/2, 0). Now here's the key observation: the zero of the polynomial 2x + 3 is –3/2, and this is exactly the x-coordinate of the point where the graph of y = 2x + 3 intersects the x-axis. Students, this is not a coincidence. In general, for any linear polynomial ax + b (with a ≠ 0), the graph of y = ax + b is a straight line that intersects the x-axis at exactly one point, namely (–b/a, 0). Therefore, the linear polynomial ax + b has exactly one zero, which is the x-coordinate of the point where the graph intersects the x-axis.

Now let's understand the geometrical meaning of zeroes of a quadratic polynomial. Consider the quadratic polynomial x² – 3x – 4. Let me create a table of values for this polynomial. When x = –2, y = (–2)² – 3 × (–2) – 4 = 4 + 6 – 4 = 6. When x = –1, y = (–1)² – 3 × (–1) – 4 = 1 + 3 – 4 = 0. When x = 0, y = 0² – 3 × 0 – 4 = –4. When x = 1, y = 1² – 3 × 1 – 4 = 1 – 3 – 4 = –6. When x = 2, y = 2² – 3 × 2 – 4 = 4 – 6 – 4 = –6. When x = 3, y = 3² – 3 × 3 – 4 = 9 – 9 – 4 = –4. When x = 4, y = 4² – 3 × 4 – 4 = 16 – 12 – 4 = 0. When x = 5, y = 5² – 3 × 5 – 4 = 25 – 15 – 4 = 6.

Now if we plot these points on a graph paper and draw the curve, we get a shape called a parabola. Students, for any quadratic polynomial ax² + bx + c where a ≠ 0, the graph of y = ax² + bx + c has one of two shapes: either it opens upwards like a cup (∪) if a > 0, or it opens downwards like an upside-down cup (∩) if a < 0. These curves are called parabolas.

Now look at our table: we found that when x = –1 and x = 4, the value of y is 0. This means the graph intersects the x-axis at two points: one at x = –1 and one at x = 4. These x-coordinates (–1 and 4) are exactly the zeroes of the quadratic polynomial x² – 3x – 4. Students, this is a very important observation: the zeroes of a quadratic polynomial are precisely the x-coordinates of the points where the parabola (the graph of y = ax² + bx + c) intersects the x-axis.

Now based on the position of the parabola relative to the x-axis, we can have three cases. Let me explain each case carefully.

In Case (i), the parabola cuts the x-axis at two distinct points. This means the quadratic polynomial has two different zeroes. The x-coordinates of these two points are the two zeroes of the polynomial.

In Case (ii), the parabola touches the x-axis at exactly one point. This happens when the two points from Case (i) coincide. In this case, the quadratic polynomial has two equal zeroes, which we can think of as one zero (but technically it's a double root). The x-coordinate of this point is the only zero for the quadratic polynomial in this case.

In Case (iii), the parabola is either completely above the x-axis or completely below the x-axis. It does not touch or cut the x-axis at all. In this case, the quadratic polynomial has no zeroes at all. Students, this is very interesting: a quadratic polynomial can have either two distinct zeroes, or two equal zeroes (which we can think of as one zero), or no zeroes at all. This also means that a polynomial of degree 2 has at most two zeroes.

Now let's think about cubic polynomials. What do you think is the geometrical meaning of zeroes of a cubic polynomial? Let's find out. Consider the cubic polynomial x³ – 4x. Let me make a table of values. When x = –2, y = (–2)³ – 4 × (–2) = –8 + 8 = 0. When x = –1, y = (–1)³ – 4 × (–1) = –1 + 4 = 3. When x = 0, y = 0³ – 4 × 0 = 0. When x = 1, y = 1³ – 4 × 1 = 1 – 4 = –3. When x = 2, y = 2³ – 4 × 2 = 8 – 8 = 0.

Now if we plot these points and draw the curve, we see that the graph intersects the x-axis at three points: x = –2, x = 0, and x = 2. These are exactly the zeroes of the cubic polynomial x³ – 4x. Students, you can see that the zeroes of a cubic polynomial are the x-coordinates of the points where the graph of y = p(x) intersects the x-axis.

Let me give you a few more examples. Consider the cubic polynomial x³. When does x³ = 0? Only when x = 0. So 0 is the only zero of x³. And if you draw the graph of y = x³, you will see that it passes through the origin (0, 0) and touches the x-axis only at that point. Now consider the polynomial x³ – x². We can factor this as x²(x – 1). So the zeroes are x = 0 (which appears twice, because we have x²) and x = 1. The graph of y = x³ – x² intersects the x-axis at x = 0 and x = 1.

From all these examples, students, we can conclude that a cubic polynomial (a polynomial of degree 3) can have at most three zeroes. In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x-axis at at most n points. Therefore, a polynomial p(x) of degree n has at most n zeroes. This is a very important result that you should remember.

Now let's look at an example from your textbook. Example 1 asks us to look at some graphs and find the number of zeroes of p(x) in each case. For graph (i), the number of zeroes is 1 because the graph intersects the x-axis at one point only. For graph (ii), the number of zeroes is 2 because the graph intersects the x-axis at two points. For graph (iii), the number of zeroes is 3. For graph (iv), the number of zeroes is 1. For graph (v), the number of zeroes is 1. For graph (vi), the number of zeroes is 4. Students, notice that in each case, the number of zeroes is exactly the number of times the graph intersects the x-axis.

Now let's move on to Section 2.3, which is about the relationship between zeroes and coefficients of a polynomial. This is a very powerful section because it shows us how the zeroes of a polynomial are connected to its coefficients.

We have already seen that the zero of a linear polynomial ax + b is –b/a, which is –(constant term)/(coefficient of x). Now let's explore the relationship for quadratic polynomials.

Consider the quadratic polynomial p(x) = 2x² – 8x + 6. In Class IX, you learned how to factorize quadratic polynomials by splitting the middle term. Let me do that here. We need to split –8x into two terms whose product is 12x² (because 6 × 2x² = 12x²). We can write 2x² – 8x + 6 = 2x² – 6x – 2x + 6. Now factor by grouping: = 2x(x – 3) – 2(x – 3) = (2x – 2)(x – 3) = 2(x – 1)(x – 3). So the zeroes are x = 1 and x = 3. Now let's look at the relationship between these zeroes and the coefficients. The sum of zeroes is 1 + 3 = 4. Notice that 4 = –(–8)/2 = –(coefficient of x)/(coefficient of x²). The product of zeroes is 1 × 3 = 3, and 3 = 6/2 = constant term/coefficient of x². So students, for this polynomial, we have sum of zeroes = –b/a and product of zeroes = c/a.

Let me take another example to verify this. Consider p(x) = 3x² + 5x – 2. We need to split 5x into two terms whose product is –6x² (because 3x² × (–2) = –6x²). We write 3x² + 5x – 2 = 3x² + 6x – x – 2 = 3x(x + 2) – 1(x + 2) = (3x – 1)(x + 2). So the zeroes are x = 1/3 and x = –2. Now sum of zeroes = 1/3 + (–2) = 1/3 – 2 = 1/3 – 6/3 = –5/3. And –(coefficient of x)/(coefficient of x²) = –5/3. So sum of zeroes = –b/a. Product of zeroes = (1/3) × (–2) = –2/3, and constant term/coefficient of x² = (–2)/3 = –2/3. So product of zeroes = c/a. This confirms our observation.

Now let's generalize this for any quadratic polynomial. Let α and β be the zeroes of the quadratic polynomial p(x) = ax² + bx + c, where a ≠ 0. Then you know that x – α and x – β are the factors of p(x). Therefore, we can write ax² + bx + c = k(x – α)(x – β), where k is some constant. Expanding the right side, we get k[x² – (α + β)x + αβ] = kx² – k(α + β)x + kαβ. Now comparing the coefficients of x², x, and constant terms on both sides, we get a = k, b = –k(α + β), and c = kαβ. From these, we can derive that α + β = –b/a and αβ = c/a. So students, here are the two important formulas: sum of zeroes = α + β = –b/a = –(coefficient of x)/(coefficient of x²), and product of zeroes = αβ = c/a = constant term/coefficient of x². These formulas are extremely useful and you should memorize them.

Now let's look at some examples from the textbook to practice these formulas.

Example 2: Find the zeroes of the quadratic polynomial x² + 7x + 10, and verify the relationship between the zeroes and the coefficients.

Solution: We have x² + 7x + 10. We need to find two numbers whose sum is 7 and product is 10. Those numbers are 2 and 5. So x² + 7x + 10 = (x + 2)(x + 5). Therefore, the zeroes are x = –2 and x = –5. Now sum of zeroes = –2 + (–5) = –7 = –(7)/1 = –(coefficient of x)/(coefficient of x²). Product of zeroes = (–2) × (–5) = 10 = 10/1 = constant term/coefficient of x². This verifies the relationship.

Example 3: Find the zeroes of the polynomial x² – 3 and verify the relationship between the zeroes and the coefficients.

Solution: Recall the identity a² – b² = (a – b)(a + b). Using this, we can write x² – 3 = x² – (√3)² = (x – √3)(x + √3). So the zeroes are x = √3 and x = –√3. Now sum of zeroes = √3 + (–√3) = 0 = –(coefficient of x)/(coefficient of x²). Since there is no x term, the coefficient of x is 0, so –0/1 = 0. Product of zeroes = √3 × (–√3) = –3 = constant term/coefficient of x² = (–3)/1 = –3. This verifies the relationship.

Example 4: Find a quadratic polynomial, the sum and product of whose zeroes are –3 and 2, respectively.

Solution: Let the quadratic polynomial be ax² + bx + c, and let its zeroes be α and β. We are given that α + β = –3 and αβ = 2. We know that α + β = –b/a and αβ = c/a. So we have –b/a = –3, which gives b/a = 3. And c/a = 2. If we take a = 1, then b = 3 and c = 2. So one quadratic polynomial that fits the given conditions is x² + 3x + 2. Students, you should know that any other quadratic polynomial that fits these conditions will be of the form k(x² + 3x + 2), where k is any real number. This is because if we multiply the polynomial by any constant k, the zeroes remain the same, but the coefficients get multiplied by k.

Now let's move on to cubic polynomials. Do you think a similar relationship exists between the zeroes of a cubic polynomial and its coefficients? Let's find out.

Consider the cubic polynomial p(x) = 2x³ – 5x² – 14x + 8. You can verify that p(x) = 0 for x = 4, x = –2, and x = 1/2. Since a cubic polynomial can have at most three zeroes, these are the zeroes of 2x³ – 5x² – 14x + 8. Now let's compute the sum of the zeroes: 4 + (–2) + 1/2 = 2 + 1/2 = 5/2. Notice that 5/2 = –(–5)/2 = –(coefficient of x²)/(coefficient of x³). Now compute the product of the zeroes: 4 × (–2) × 1/2 = –4 = –8/2 = –(constant term)/(coefficient of x³). However, there is one more relationship. Consider the sum of the products of the zeroes taken two at a time. We have (4 × (–2)) + ((–2) × 1/2) + (1/2 × 4) = –8 – 1 + 2 = –7 = –14/2 = coefficient of x/(coefficient of x³). So students, for cubic polynomials, we have three relationships.

In general, it can be proved that if α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d, then: α + β + γ = –b/a, αβ + βγ + γα = c/a, and αβγ = –d/a.

These are the three important formulas for cubic polynomials. Let me say them again so that you remember them clearly. The sum of the zeroes is equal to –(coefficient of x²)/(coefficient of x³). The sum of the products of the zeroes taken two at a time is equal to (coefficient of x)/(coefficient of x³). And the product of the zeroes is equal to –(constant term)/(coefficient of x³).

Now let's look at Example 5 from the textbook. We need to verify that 3, –1, and –1/3 are the zeroes of the cubic polynomial p(x) = 3x³ – 5x² – 11x – 3, and then verify the relationship between the zeroes and the coefficients.

Solution: First, we compare the given polynomial with ax³ + bx² + cx + d. So a = 3, b = –5, c = –11, d = –3. Now we need to verify that these numbers are zeroes. We compute p(3) = 3 × 3³ – 5 × 3² – 11 × 3 – 3 = 3 × 27 – 5 × 9 – 33 – 3 = 81 – 45 – 33 – 3 = 0. So 3 is a zero. Next, p(–1) = 3 × (–1)³ – 5 × (–1)² – 11 × (–1) – 3 = 3 × (–1) – 5 × 1 + 11 – 3 = –3 – 5 + 11 – 3 = 0. So –1 is a zero. Finally, p(–1/3) = 3 × (–1/3)³ – 5 × (–1/3)² – 11 × (–1/3) – 3 = 3 × (–1/27) – 5 × (1/9) + 11/3 – 3 = –1/9 – 5/9 + 11/3 – 3 = –6/9 + 11/3 – 3 = –2/3 + 11/3 – 3 = 9/3 – 3 = 3 – 3 = 0. So –1/3 is also a zero. Therefore, 3, –1, and –1/3 are the zeroes of 3x³ – 5x² – 11x – 3.

Now let α = 3, β = –1, and γ = –1/3. Let's verify the relationships. First, α + β + γ = 3 + (–1) + (–1/3) = 2 – 1/3 = 6/3 – 1/3 = 5/3. And –b/a = –(–5)/3 = 5/3. So sum of zeroes = –b/a. Next, αβ + βγ + γα = 3 × (–1) + (–1) × (–1/3) + (–1/3) × 3 = –3 + 1/3 – 1 = –4 + 1/3 = –12/3 + 1/3 = –11/3. And c/a = (–11)/3 = –11/3. So sum of products of zeroes taken two at a time = c/a. Finally, αβγ = 3 × (–1) × (–1/3) = 3 × 1/3 = 1. And –d/a = –(–3)/3 = 3/3 = 1. So product of zeroes = –d/a. All three relationships are verified.

Students, this is exactly what we expected from the formulas we derived earlier. The relationship between zeroes and coefficients is very elegant and useful. It allows us to find the sum, product, and other combinations of zeroes without actually finding the zeroes themselves, if we know the coefficients of the polynomial.

Now let's summarize everything we have learned in this chapter.

In this chapter, you have studied the following important points:

First, polynomials of degrees 1, 2, and 3 are called linear, quadratic, and cubic polynomials respectively. A linear polynomial has the form ax + b, a quadratic polynomial has the form ax² + bx + c, and a cubic polynomial has the form ax³ + bx² + cx + d, where in each case the leading coefficient (the coefficient of the highest power) is not zero.

Second, a quadratic polynomial in x with real coefficients is of the form ax² + bx + c, where a, b, c are real numbers with a ≠ 0.

Third, the zeroes of a polynomial p(x) are precisely the x-coordinates of the points where the graph of y = p(x) intersects the x-axis. This is the geometrical meaning of zeroes.

Fourth, a quadratic polynomial can have at most 2 zeroes, and a cubic polynomial can have at most 3 zeroes. In general, a polynomial of degree n can have at most n zeroes.

Fifth, if α and β are the zeroes of the quadratic polynomial ax² + bx + c, then the sum of the zeroes is α + β = –b/a, which is –(coefficient of x)/(coefficient of x²), and the product of the zeroes is αβ = c/a, which is constant term/(coefficient of x²).

Sixth, if α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d, then the sum of the zeroes is α + β + γ = –b/a, which is –(coefficient of x²)/(coefficient of x³). The sum of the products of the zeroes taken two at a time is αβ + βγ + γα = c/a, which is (coefficient of x)/(coefficient of x³). And the product of the zeroes is αβγ = –d/a, which is –(constant term)/(coefficient of x³).

These relationships are extremely useful in solving many problems related to polynomials. You can use them to find the sum, product, and other combinations of zeroes when you know the coefficients of the polynomial. Conversely, you can also form a polynomial when you know the sum and product of its zeroes.

Students, this brings us to the end of our lesson on polynomials. We have covered all the important concepts: what polynomials are, the different types of polynomials based on their degrees, the concept of zeroes and their geometrical meaning, and most importantly, the relationship between zeroes and coefficients for both quadratic and cubic polynomials. These concepts are fundamental to understanding algebra and will be very useful in your future studies of mathematics.

Remember, practice is the key to mastering polynomials. Make sure you solve many problems based on these concepts. Thank you for your attention, and I hope you enjoyed this lesson. Keep studying hard!