Gonit Vigyan

 

📢 Introduction: Exploring Polynomials Easily

Polynomials are fundamental in algebra, helping us understand mathematical patterns. In Class IX, you learned about their degrees, types (linear, quadratic, cubic), and zeroes. But how are zeroes connected to coefficients?

For example, in p(x) = x² – 3x – 4, substituting x = –1 gives p(–1) = 0, meaning –1 is a zero. But what does this tell us?

In this post, we'll explore zeroes of polynomials, their link to coefficients, and the division algorithm, making these concepts simple and clear. Let’s dive in! 🚀

📖 Polynomials: General Form, Types, and Zeroes

📌 General Equation of a Polynomial

A polynomial in one variable $x$ is expressed as:

$$p(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_2 x^2 + a_1 x + a_0$$

where:

• $n$ is a non-negative integer (degree of the polynomial).
• $a_n, a_{n-1}, \dots, a_1, a_0$ are real coefficients.
• $a_n \neq 0$ (leading coefficient).

---

📌 Types of Polynomials & Their General Forms

1️⃣ Linear Polynomial (Degree = 1)

$$p(x) = ax + b, \quad a \neq 0$$

✅ Example: $3x + 5$, $-2y + 7$

2️⃣ Quadratic Polynomial (Degree = 2)

$$p(x) = ax^2 + bx + c, \quad a \neq 0$$

✅ Example: $x^2 - 3x + 2$, $2y^2 + 5y - 7$

3️⃣ Cubic Polynomial (Degree = 3)

$$p(x) = ax^3 + bx^2 + cx + d, \quad a \neq 0$$

✅ Example: $x^3 - 4x^2 + 2x - 1$, $3y^3 + 7y^2 - 2y + 5$

4️⃣ Higher-Degree Polynomials (Degree $n$)

$$p(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$$

✅ Example (Degree 6): $7x^6 - 3x^4 + 2x^2 - 8$

---

📌 Evaluating a Polynomial at a Given Value

• The value of a polynomial at $x = k$ is found by substituting $x$ with $k$.
• Example: If $p(x) = x^2 - 3x - 4$, then:
  • $p(2) = 2^2 - 3(2) - 4 = 4 - 6 - 4 = -6$
  • $p(0) = 0^2 - 3(0) - 4 = -4$

---

📌 Zeroes of a Polynomial

• A real number $k$ is a zero of a polynomial $p(x)$ if $p(k) = 0$.
• Example: For $p(x) = x^2 - 3x - 4$:
  • $p(-1) = (-1)^2 - 3(-1) - 4 = 0$
  • $p(4) = 4^2 - 3(4) - 4 = 0$
• So, $-1$ and $4$ are the zeroes of $x^2 - 3x - 4$.

🔹 Zeroes of a Linear Polynomial

• A linear polynomial $p(x) = ax + b$ has only one zero, given by: $$k = \frac{-b}{a}$$
• Example: Find the zero of $p(x) = 2x + 3$: $$2x + 3 = 0 \Rightarrow x = -\frac{3}{2}$$
• General Formula: $$\text{Zero of } ax + b = \frac{-\text{(constant term)}}{\text{coefficient of } x}$$

🔹 Key Question: Do quadratic polynomials follow a similar pattern? 🤔

---

📌 Summary

✅ A polynomial is an algebraic expression with terms having non-negative exponents.
✅ Types of polynomials:

• Linear (Degree 1): $ax + b$
• Quadratic (Degree 2): $ax^2 + bx + c$
• Cubic (Degree 3): $ax^3 + bx^2 + cx + d$
  ✅ Zeroes of polynomials:
• Linear Polynomial: Zero = $-\frac{b}{a}$.
• Quadratic Polynomial: Zeroes satisfy $p(x) = 0$.
  ✅ Next, we explore how zeroes of a quadratic polynomial relate to its coefficients and the division algorithm for polynomials. 🚀

How to Identify Whether an Expression is a Polynomial or Not?

A polynomial is an algebraic expression consisting of variables, coefficients, and exponents. However, not all algebraic expressions are polynomials.

📌 Rules for an Expression to Be a Polynomial

✅ The exponents of the variables must be whole numbers (0, 1, 2, 3, …).
✅ The coefficients must be real numbers.
✅ No variables in the denominator.
✅ No negative or fractional exponents.
✅ No square roots or other irrational functions of variables.

---

📌 Examples of Polynomials

Expression	Polynomial?	Reason
$4x + 2$	✅ Yes	Degree is 1 (whole number exponent).
$5x^3 - 4x^2 + x - 2$	✅ Yes	Highest power is 3 (whole number exponent).
$7u^6 - 3u^4 + 2u^2 - 8$	✅ Yes	Exponents are whole numbers.
$x^2 - 3x - 4$	✅ Yes	Quadratic polynomial (degree 2).

---

📌 Examples of Non-Polynomials

Expression	Polynomial?	Reason
$\frac{1}{x} - 1$	❌ No	$x^{-1}$ (negative exponent).
$x + 2$	✅ Yes	Linear polynomial (degree 1).
$\frac{1}{x^2} + 2x + 3$	❌ No	$x^{-2}$ (negative exponent).
$\sqrt{x} + 3x - 1$	❌ No	$x^{1/2}$ (fractional exponent).
$2^x + x^3$	❌ No	Exponent is a variable in $2^x$.

---

📌 Quick Checklist to Identify a Polynomial

✅ Whole number exponents only (e.g., 0, 1, 2, 3, …).
✅ No variables in denominators (e.g., $\frac{1}{x}$ is not allowed).
✅ No negative or fractional exponents (e.g., $x^{-1}$, $x^{1/2}$).
✅ No square roots or trigonometric functions (e.g., $\sqrt{x}$, $\sin x$).

---

💡 Tip: If an expression follows these rules, it is a polynomial. If it violates any, it is not a polynomial. 🚀

📖 Geometrical Meaning of the Zeroes of a Polynomial

📌 What Are Zeroes of a Polynomial?

A real number $k$ is a zero of a polynomial $p(x)$ if:

$$p(k) = 0$$

But why are zeroes important? To understand this, we examine the geometrical interpretation of zeroes using graphs of polynomials.

---

📌 Zeroes of a Linear Polynomial

A linear polynomial is of the form:

$$p(x) = ax + b, \quad a \neq 0$$

• The graph of $y = ax + b$ is a straight line.
• It intersects the x-axis at one point, where $y = 0$.
• The zero of the polynomial is the x-coordinate of this intersection point.

Example:

For $y = 2x + 3$, the graph intersects the x-axis at:

$$x = -\frac{3}{2}$$

Thus, $-\frac{3}{2}$ is the zero of $p(x) = 2x + 3$.

✅ Conclusion: A linear polynomial always has exactly one zero.

---

📌 Zeroes of a Quadratic Polynomial

A quadratic polynomial is of the form:

$$p(x) = ax^2 + bx + c, \quad a \neq 0$$

• The graph of $y = ax^2 + bx + c$ is a parabola (U-shaped curve).
• The zeroes of the polynomial are the x-coordinates of the points where the parabola intersects the x-axis.

Example:

Consider $p(x) = x^2 - 3x - 4$.
From calculations, its zeroes are $x = -1$ and $x = 4$.

• The graph of $y = x^2 - 3x - 4$ intersects the x-axis at $(-1, 0)$ and $(4,0)$.
• Hence, -1 and 4 are the zeroes of the quadratic polynomial.

✅ Conclusion: A quadratic polynomial can have at most two zeroes.

---

📌 Three Possible Cases for a Quadratic Polynomial

Depending on the number of intersections with the x-axis, there are three cases:

1️⃣ Two Distinct Zeroes (Graph intersects at two points)

• The equation $ax^2 + bx + c = 0$ has two distinct real roots.
• Example: $y = x^2 - 3x - 4$

2️⃣ One Zero (Double Root) (Graph touches the x-axis at one point)

• The equation $ax^2 + bx + c = 0$ has one repeated root.
• Example: $y = (x - 2)^2$, which touches the x-axis at $x = 2$.

3️⃣ No Real Zeroes (Graph does not touch the x-axis)

• The equation $ax^2 + bx + c = 0$ has no real solutions.
• Example: $y = x^2 + 4$, which is always positive and never crosses the x-axis.

✅ Conclusion: A quadratic polynomial has either 2, 1, or 0 real zeroes.

---

📌 Zeroes of a Cubic Polynomial

A cubic polynomial is of the form:

$$p(x) = ax^3 + bx^2 + cx + d, \quad a \neq 0$$

• The graph of $y = ax^3 + bx^2 + cx + d$ is a curve that can intersect the x-axis at most three times.
• The zeroes of the polynomial are the x-coordinates of these intersection points.

Example:

Consider $p(x) = x^3 - 4x$.

• The graph intersects the x-axis at $x = -2, 0, 2$.
• So, the zeroes are -2, 0, and 2.

✅ Conclusion: A cubic polynomial can have at most three zeroes.

---

📌 General Rule: Number of Zeroes and Degree of the Polynomial

• A polynomial of degree $n$ can have at most $n$ zeroes.
• The graph of $y = p(x)$ intersects the x-axis at most $n$ times.

Degree	Maximum Number of Zeroes	Example
1 (Linear)	1	$y = 2x + 3$
2 (Quadratic)	2	$y = x^2 - 3x - 4$
3 (Cubic)	3	$y = x^3 - 4x$
4 (Quartic)	4	$y = x^4 - 2x^2 + 1$
$n$ (General)	$n$	$y = x^n - 1$

---

📌 Summary

✅ The zeroes of a polynomial are the x-coordinates where its graph intersects the x-axis.
✅ A linear polynomial has exactly one zero.
✅ A quadratic polynomial has at most two zeroes.
✅ A cubic polynomial has at most three zeroes.
✅ In general, a polynomial of degree $n$ has at most $n$ zeroes.

🚀 Understanding these concepts helps in solving polynomial equations graphically and algebraically!

Let's visualize these three cases of quadratic polynomials by plotting their graphs:

1️⃣ Case (i): The graph intersects the x-axis at two distinct points → Two distinct zeroes.

• Example: $y = x^2 - 3x - 4$

2️⃣ Case (ii): The graph touches the x-axis at exactly one point → One repeated zero.

• Example: $y = (x - 2)^2$

3️⃣ Case (iii): The graph does not intersect the x-axis → No real zeroes.

• Example: $y = x^2 + 4$

Now, let's plot these cases. 🎨📊

Here are the graphs illustrating the three cases of quadratic polynomials:

1️⃣ Case (i): Two Distinct Zeroes → The graph intersects the x-axis at two points.

• Example: $y = x^2 - 3x - 4$.

2️⃣ Case (ii): One Repeated Zero → The graph touches the x-axis at one point.

• Example: $y = (x - 2)^2$.

3️⃣ Case (iii): No Real Zeroes → The graph does not intersect the x-axis.

• Example: $y = x^2 + 4$.

These cases visually explain how quadratic equations can have two, one, or zero real solutions depending on their graph. 🚀📊

I can't directly insert the image into the response, but I can format the question and include a reference to the provided graph.

---

EXERCISE 2.1

Question 1:

The graphs of $y = p(x)$ are given in Fig. 2.10 below, for some polynomials $p(x)$. Find the number of zeroes of $p(x)$ in each case.

   

---

Answer:

(i)

• Number of Zeroes: 0
• Reason: The graph does not intersect the x-axis.

(ii)

• Number of Zeroes: 1
• Reason: The graph touches the x-axis at one point.

(iii)

• Number of Zeroes: 3
• Reason: The graph crosses the x-axis three times.

(iv)

• Number of Zeroes: 1
• Reason: The graph touches the x-axis at exactly one point (double root).

(v)

• Number of Zeroes: 2
• Reason: The graph crosses the x-axis twice.

(vi)

• Number of Zeroes: 4
• Reason: The graph crosses the x-axis four times.

📖 Understanding the Relationship Between Zeroes and Coefficients of a Polynomial

We’ve already seen that the zero of a linear polynomial $ax + b$ is found using the formula:

$$x = -\frac{b}{a}$$

But what about quadratic polynomials? Is there a pattern between their zeroes and coefficients? Let’s explore this step by step.

---

🔹 Example 1: Finding the Zeroes of a Quadratic Polynomial

Let’s take the quadratic polynomial:

$$p(x) = 2x^2 - 8x + 6$$

We’ll factorize it using the splitting the middle term method:

1️⃣ Multiply the coefficient of $x^2$ (which is 2) and the constant term (6):

$$2 \times 6 = 12$$

2️⃣ Find two numbers that multiply to 12 and add up to -8 (the coefficient of $x$):

• The numbers -6 and -2 work because: $$(-6) + (-2) = -8, \quad (-6) \times (-2) = 12$$

3️⃣ Rewrite the middle term (-8x) using these numbers:

$$2x^2 - 6x - 2x + 6$$

4️⃣ Factor in pairs:

$$2x(x - 3) - 2(x - 3)$$

5️⃣ Take out the common factor:

$$(2x - 2)(x - 3) = 2(x - 1)(x - 3)$$

Now, setting $p(x) = 0$:

$$2(x - 1)(x - 3) = 0$$

So,

$$x - 1 = 0 \quad \text{or} \quad x - 3 = 0$$ $$x = 1, \quad x = 3$$

✅ The zeroes of the polynomial are 1 and 3.

---

🔹 Observing the Relationship

Now, let’s check if there’s a pattern:

• Sum of zeroes:

  $$1 + 3 = 4$$

  This matches the formula:

  $$-\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{-8}{2} = 4$$

• Product of zeroes:

  $$1 \times 3 = 3$$

  This matches:

  $$\frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{6}{2} = 3$$

🎯 Conclusion: The sum and product of zeroes are directly related to the coefficients of the polynomial!

---

🔹 Example 2: Another Quadratic Polynomial

Let’s try another example:

$$p(x) = 3x^2 + 5x - 2$$

Step-by-step Factorization

1️⃣ Multiply the coefficient of $x^2$ (3) and the constant term (-2):

$$3 \times (-2) = -6$$

2️⃣ Find two numbers that multiply to -6 and add up to 5:

• The numbers 6 and -1 work because: $$6 + (-1) = 5, \quad 6 \times (-1) = -6$$

3️⃣ Rewrite the middle term (5x) using these numbers:

$$3x^2 + 6x - x - 2$$

4️⃣ Factor in pairs:

$$3x(x + 2) - 1(x + 2)$$

5️⃣ Take out the common factor:

$$(3x - 1)(x + 2)$$

Now, setting $p(x) = 0$:

$$3x - 1 = 0 \quad \text{or} \quad x + 2 = 0$$ $$x = \frac{1}{3}, \quad x = -2$$

✅ The zeroes are $\frac{1}{3}$ and -2.

Now, let’s check the sum and product:

• Sum of zeroes:

  $$\frac{1}{3} + (-2) = -\frac{5}{3}$$

  This matches:

  $$-\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{5}{3}$$

• Product of zeroes:

  $$\frac{1}{3} \times (-2) = -\frac{2}{3}$$

  This matches:

  $$\frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{-2}{3}$$

---

🔹 General Formula for a Quadratic Polynomial

For any quadratic polynomial:

$$p(x) = ax^2 + bx + c, \quad a \neq 0$$

Let $\alpha$ and $\beta$ be the two zeroes.

🔹 Sum of zeroes:

$$\alpha + \beta = -\frac{b}{a}$$

🔹 Product of zeroes:

$$\alpha \beta = \frac{c}{a}$$

---

🔹 Summary Table

Polynomial	Zeroes	Sum of Zeroes $(\alpha + \beta)$	Product of Zeroes $(\alpha \beta)$
$2x^2 - 8x + 6$	$1, 3$	$4$	$3$
$3x^2 + 5x - 2$	$\frac{1}{3}, -2$	$-\frac{5}{3}$	$-\frac{2}{3}$

---

🔹 Key Takeaways

✔️ The sum of zeroes is always given by:

$$-\frac{\text{coefficient of } x}{\text{coefficient of } x^2}$$

✔️ The product of zeroes is always given by:

$$\frac{\text{constant term}}{\text{coefficient of } x^2}$$

✔️ This pattern holds for every quadratic polynomial!

 

📖 Understanding the Relationship Between Zeroes and Coefficients of a Polynomial

Polynomials are essential in algebra, and their zeroes tell us a lot about their structure. In this section, we explore how the zeroes of a polynomial relate to its coefficients using simple mathematical observations.

---

📌 Zero of a Linear Polynomial

A linear polynomial is of the form:

$$p(x) = ax + b, \quad a \neq 0$$

The zero of the polynomial is found by solving $p(x) = 0$:

$$ax + b = 0 \quad \Rightarrow \quad x = -\frac{b}{a}$$

✅ Conclusion: The zero of a linear polynomial is simply $-\frac{\text{constant term}}{\text{coefficient of } x}$.

---

📌 Relationship Between Zeroes and Coefficients of a Quadratic Polynomial

A quadratic polynomial is of the form:

$$p(x) = ax^2 + bx + c, \quad a \neq 0$$

The zeroes of this polynomial are the values of $x$ where $p(x) = 0$.

Example 1: Finding Zeroes and Their Relationship

Let’s take the quadratic polynomial:

$$p(x) = 2x^2 - 8x + 6$$

We factorize it:

$$2x^2 - 8x + 6 = 2(x - 1)(x - 3)$$

Setting $p(x) = 0$:

$$(x - 1) = 0 \quad \text{or} \quad (x - 3) = 0$$

Thus, the zeroes are $x = 1$ and $x = 3$.

Observations

• Sum of Zeroes: $$1 + 3 = \frac{-(-8)}{2} = \frac{\text{coefficient of } x}{\text{coefficient of } x^2}$$
• Product of Zeroes: $$1 \times 3 = \frac{6}{2} = \frac{\text{constant term}}{\text{coefficient of } x^2}$$

✅ Conclusion: For any quadratic polynomial $ax^2 + bx + c$:

$$\text{Sum of zeroes} = -\frac{b}{a}, \quad \text{Product of zeroes} = \frac{c}{a}$$

---

📌 Verifying with Another Example

Consider another quadratic polynomial:

$$p(x) = 3x^2 + 5x - 2$$

Factorizing:

$$p(x) = (3x - 1)(x + 2)$$

Setting $p(x) = 0$:

$$3x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{3}, \quad \text{or} \quad x + 2 = 0 \quad \Rightarrow \quad x = -2$$

Thus, the zeroes are $\frac{1}{3}$ and $-2$.

Verifying the Relationship

• Sum of Zeroes: $$\frac{1}{3} + (-2) = -\frac{5}{3} = -\frac{b}{a}$$
• Product of Zeroes: $$\frac{1}{3} \times (-2) = -\frac{2}{3} = \frac{c}{a}$$

Thus, the relationships hold true again! ✅

---

📌 General Rule for a Quadratic Polynomial

For any quadratic polynomial $ax^2 + bx + c$:

$$\text{Sum of zeroes} = -\frac{b}{a}, \quad \text{Product of zeroes} = \frac{c}{a}$$

This formula helps find missing coefficients or zeroes without factorization.

---

📌 Finding a Quadratic Polynomial from Given Zeroes

If we know the sum and product of the zeroes, we can construct the quadratic polynomial.

Example 3: Find the Quadratic Polynomial

If the sum of zeroes is -3 and the product of zeroes is 2, then:

$$p(x) = x^2 - (\text{sum of zeroes}) x + (\text{product of zeroes})$$

Substituting values:

$$p(x) = x^2 + 3x + 2$$

Thus, any polynomial of the form $k(x^2 + 3x + 2)$ (where $k$ is a constant) will satisfy this condition.

---

📌 Relationship Between Zeroes and Coefficients of a Cubic Polynomial

A cubic polynomial is of the form:

$$p(x) = ax^3 + bx^2 + cx + d$$

If $\alpha, \beta, \gamma$ are its zeroes, then:

• Sum of Zeroes: $$\alpha + \beta + \gamma = -\frac{b}{a}$$
• Sum of Products of Zeroes (Taken Two at a Time): $$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$$
• Product of Zeroes: $$\alpha\beta\gamma = -\frac{d}{a}$$

---

📌 Example: Finding Zeroes of a Cubic Polynomial

Consider:

$$p(x) = 2x^3 - 5x^2 - 14x + 8$$

We find that the zeroes are $4, -2, \frac{1}{2}$.

Verifying the Relationship

• Sum of Zeroes: $$4 + (-2) + \frac{1}{2} = -\frac{-5}{2} = -\frac{b}{a}$$
• Sum of Products of Zeroes (Taken Two at a Time): $$(4 \times -2) + (-2 \times \frac{1}{2}) + (4 \times \frac{1}{2}) = \frac{c}{a}$$
• Product of Zeroes: $$4 \times (-2) \times \frac{1}{2} = -\frac{8}{2} = -\frac{d}{a}$$

Thus, the relationships hold true again! ✅

---

📌 Summary of Key Formulas

Polynomial Type	Sum of Zeroes	Product of Zeroes	Extra Relationship
Linear $ax + b$	$-\frac{b}{a}$	N/A	N/A
Quadratic $ax^2 + bx + c$	$-\frac{b}{a}$	$\frac{c}{a}$	N/A
Cubic $ax^3 + bx^2 + cx + d$	$-\frac{b}{a}$	$-\frac{d}{a}$	$\frac{c}{a}$ (Sum of product of pairs)

---

🚀 Why Is This Useful?

• Helps in finding unknown coefficients in polynomials.
• Allows us to quickly form polynomials when given the zeroes.
• Essential for higher-level algebra and real-world applications.

Mastering these relationships makes solving polynomial equations faster and easier! 💡✨

📖 EXERCISE 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) $x^2 - 2x - 8$
(ii) $4s^2 - 4s + 1$
(iii) $6x^2 - 3 - 7x$
(iv) $4u^2 + 8u$
(v) $t^2 - 15$
(vi) $3x^2 - x - 4$

---

Solution 1: Finding Zeroes and Verifying Relationship

(i) $x^2 - 2x - 8$

Factorizing using middle-term split:

$$x^2 - 4x + 2x - 8 = x(x - 4) + 2(x - 4)$$ $$(x + 2)(x - 4) = 0$$

Setting each factor to zero:

$$x + 2 = 0 \Rightarrow x = -2, \quad x - 4 = 0 \Rightarrow x = 4$$

Zeroes: $-2$ and $4$.

Verification:

• Sum of Zeroes: $(-2) + 4 = 2 = -\frac{-2}{1}$ ✅
• Product of Zeroes: $(-2) \times 4 = -8 = \frac{-8}{1}$ ✅

---

(ii) $4s^2 - 4s + 1$

This is a perfect square trinomial:

$$(2s - 1)(2s - 1) = 0$$ $$(2s - 1)^2 = 0$$

Setting $2s - 1 = 0$:

$$2s = 1 \Rightarrow s = \frac{1}{2}$$

Zeroes: $\frac{1}{2}, \frac{1}{2}$ (repeated).

Verification:

• Sum of Zeroes: $\frac{1}{2} + \frac{1}{2} = 1 = -\frac{-4}{4}$ ✅
• Product of Zeroes: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{1}{4}$ ✅

---

(iii) $6x^2 - 3 - 7x$

Rearrange:

$$6x^2 - 7x - 3 = 0$$

Split middle term:

$$6x^2 - 9x + 2x - 3 = 0$$

Factorize:

$$3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3) = 0$$

Setting each factor to zero:

$$3x + 1 = 0 \Rightarrow x = -\frac{1}{3}, \quad 2x - 3 = 0 \Rightarrow x = \frac{3}{2}$$

Zeroes: $-\frac{1}{3}, \frac{3}{2}$.

Verification:

• Sum of Zeroes: $-\frac{1}{3} + \frac{3}{2} = \frac{-2 + 9}{6} = \frac{7}{6} = -\frac{-7}{6}$ ✅
• Product of Zeroes: $-\frac{1}{3} \times \frac{3}{2} = -\frac{3}{6} = -\frac{1}{2} = \frac{-3}{6}$ ✅

---

(iv) $4u^2 + 8u$

Factor out common term:

$$4u(u + 2) = 0$$

Setting each factor to zero:

$$4u = 0 \Rightarrow u = 0, \quad u + 2 = 0 \Rightarrow u = -2$$

Zeroes: $0, -2$.

Verification:

• Sum of Zeroes: $0 + (-2) = -2 = -\frac{8}{4}$ ✅
• Product of Zeroes: $0 \times (-2) = 0 = \frac{0}{4}$ ✅

---

(v) $t^2 - 15$

Using identity:

$$t^2 - 15 = (t - \sqrt{15})(t + \sqrt{15}) = 0$$

Setting each factor to zero:

$$t - \sqrt{15} = 0 \Rightarrow t = \sqrt{15}, \quad t + \sqrt{15} = 0 \Rightarrow t = -\sqrt{15}$$

Zeroes: $\sqrt{15}, -\sqrt{15}$.

Verification:

• Sum of Zeroes: $\sqrt{15} + (-\sqrt{15}) = 0 = -\frac{0}{1}$ ✅
• Product of Zeroes: $\sqrt{15} \times (-\sqrt{15}) = -15 = \frac{-15}{1}$ ✅

---

(vi) $3x^2 - x - 4$

Splitting the middle term:

$$3x^2 - 4x + 3x - 4 = 0$$

Factorize:

$$x(3x + 3) - 4(3x + 3) = (x - 4)(3x + 3) = 0$$

Setting each factor to zero:

$$x - 4 = 0 \Rightarrow x = 4, \quad 3x + 3 = 0 \Rightarrow x = -1$$

Zeroes: $4, -1$.

Verification:

• Sum of Zeroes: $4 + (-1) = 3 = -\frac{-1}{3}$ ✅
• Product of Zeroes: $4 \times (-1) = -4 = \frac{-4}{3}$ ✅

---

2. Find a quadratic polynomial for the given sum and product of zeroes.

Using the formula:

$$p(x) = x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})$$

(i) $\frac{1}{4}, -1$

$$p(x) = x^2 - \left(\frac{1}{4} - 1\right)x + \left(\frac{1}{4} \times (-1)\right)$$ $$p(x) = x^2 - \left(\frac{1 - 4}{4}\right)x + \left(-\frac{1}{4}\right)$$ $$p(x) = x^2 - \left(-\frac{3}{4}\right)x - \frac{1}{4}$$ $$p(x) = 4x^2 + 3x - 1$$

(ii) $\frac{1}{2}, \frac{3}{4}$

$$p(x) = x^2 - \frac{1}{2}x + \frac{3}{4}$$

Multiplying by 4 to remove fractions:

$$4x^2 - 2x + 3$$

(iii) $0, 5$

$$p(x) = x^2 - (0 + 5)x + (0 \times 5)$$ $$p(x) = x^2 - 5x$$

---

 

📖 Notes on Polynomial Division & Finding Zeroes

1️⃣ Understanding the Division Algorithm for Polynomials

The Division Algorithm for Polynomials states that for any two polynomials $p(x)$ and $g(x)$ (where $g(x) \neq 0$), there exist unique polynomials $q(x)$ (quotient) and $r(x)$ (remainder) such that:

$$p(x) = g(x) \times q(x) + r(x)$$

where the degree of $r(x)$ is less than the degree of $g(x)$, or $r(x) = 0$.

This algorithm helps in:
✅ Finding unknown zeroes of a polynomial when some zeroes are already known.
✅ Checking if a given polynomial is a factor of another polynomial.

---

2️⃣ Finding the Remaining Zeroes of a Cubic Polynomial

Example 1: Find the remaining zeroes of $x^3 - 3x^2 - x + 3$, given that one zero is 1.

✅ Step 1: Given information

• The polynomial is $x^3 - 3x^2 - x + 3$
• Given that one of its zeroes is 1, so $x - 1$ is a factor.

✅ Step 2: Perform Polynomial Division

$$\begin{array}{r|rrrr} & x^2 - 2x - 3 \\ \hline x - 1 \, & x^3 - 3x^2 - x + 3 \\ & \underline{x^3 - x^2} \\ \hline & -2x^2 - x + 3 \\ & \underline{-2x^2 + 2x} \\ \hline & -3x + 3 \\ & \underline{-3x + 3} \\ \hline & 0 \\ \end{array}$$

✅ Quotient: $x^2 - 2x - 3$
✅ Remainder: $0$

✅ Step 3: Factorizing the Quotient

$$x^2 - 2x - 3 = (x + 1)(x - 3)$$

So,

$$x^3 - 3x^2 - x + 3 = (x - 1)(x + 1)(x - 3)$$

✅ Final Zeroes: $1, -1, 3$

---

3️⃣ Polynomial Division of a Quadratic Polynomial

Example 2: Divide $2x^2 + 3x + 1$ by $x + 2$

✅ Step 1: Polynomial Division

$$\begin{array}{r|rrrr} & 2x - 1 \\ \hline x + 2 \, & 2x^2 + 3x + 1 \\ & \underline{2x^2 + 4x} \\ \hline & -x + 1 \\ & \underline{-x - 2} \\ \hline & 3 \\ \end{array}$$

✅ Quotient: $2x - 1$
✅ Remainder: $3$

$$2x^2 + 3x + 1 = (x + 2)(2x - 1) + 3$$

---

4️⃣ Polynomial Division by a Quadratic Polynomial

Example 3: Divide $3x^3 + x^2 + 2x + 5$ by $x^2 + 2x + 1$

✅ Step 1: Polynomial Division

$$\begin{array}{r|rrrr} & 3x - 5 \\ \hline x^2 + 2x + 1 \, & 3x^3 + x^2 + 2x + 5 \\ & \underline{3x^3 + 6x^2 + 3x} \\ \hline & -5x^2 - x + 5 \\ & \underline{-5x^2 - 10x - 5} \\ \hline & 9x + 10 \\ \end{array}$$

✅ Quotient: $3x - 5$
✅ Remainder: $9x + 10$

$$3x^3 + x^2 + 2x + 5 = (x^2 + 2x + 1)(3x - 5) + (9x + 10)$$

---

5️⃣ Finding All Zeroes of a Quartic Polynomial

Example 4: Find all zeroes of $2x^4 - 3x^3 - 3x^2 + 6x - 2$, given that two zeroes are $2$ and $-2$.

✅ Step 1: Given information
Since $2$ and $-2$ are zeroes,

$$(x - 2)(x + 2) = x^2 - 4$$

is a factor of the polynomial.

✅ Step 2: Perform Polynomial Division

$$\begin{array}{r|rrrr} & 2x^2 - 3x + 1 \\ \hline x^2 - 2 \, & 2x^4 - 3x^3 - 3x^2 + 6x - 2 \\ & \underline{2x^4 - 4x^2} \\ \hline & -3x^3 + x^2 + 6x - 2 \\ & \underline{-3x^3 + 6x} \\ \hline & x^2 - 2 \\ & \underline{x^2 - 2} \\ \hline & 0 \\ \end{array}$$

✅ Quotient: $2x^2 - 3x + 1$
✅ Remainder: $0$

✅ Step 3: Factorizing the Quotient

$$2x^2 - 3x + 1 = (2x - 1)(x - 1)$$

✅ Final Zeroes: $2, -2, \frac{1}{2}, 1$

---

6️⃣ Summary of Key Concepts

✔ Polynomial Division Algorithm:

$$p(x) = g(x) \times q(x) + r(x)$$

where $r(x) = 0$ or $\deg r(x) < \deg g(x)$.

✔ Finding Zeroes of a Polynomial:
1️⃣ Divide the polynomial by a known factor (given zero).
2️⃣ Factorize the quotient polynomial.
3️⃣ The roots of the factored form give all the zeroes.

✔ Factorization by Middle Term Splitting:
Used to break quadratic polynomials into linear factors

📖 Exercise 2.3 - Solutions & Notes on Polynomial Division

---

1️⃣ Divide the polynomial $p(x)$ by $g(x)$ and find the quotient and remainder.

(i) $p(x) = x^3 - 3x^2 + 5x - 3$, $g(x) = x^2 - 2$

✅ Step 1: Perform Polynomial Division

$$\begin{array}{r|rrrr} & x - 3 \\ \hline x^2 - 2 \, & x^3 - 3x^2 + 5x - 3 \\ & \underline{x^3 - 2x} \\ \hline & -3x^2 + 7x - 3 \\ & \underline{-3x^2 + 6} \\ \hline & 7x - 9 \\ \end{array}$$

✅ Quotient: $x - 3$
✅ Remainder: $7x - 9$

---

(ii) $p(x) = x^4 - 3x^2 + 4x + 5$, $g(x) = x^2 + 1 - x$

✅ Rewriting the divisor in standard form

$$g(x) = x^2 - x + 1$$

✅ Performing Polynomial Division

$$\begin{array}{r|rrrr} & x^2 + x - 2 \\ \hline x^2 - x + 1 \, & x^4 - 3x^2 + 4x + 5 \\ & \underline{x^4 - x^3 + x^2} \\ \hline & x^3 - 4x^2 + 4x + 5 \\ & \underline{x^3 - x^2 + x} \\ \hline & -3x^2 + 3x + 5 \\ & \underline{-3x^2 + 3x - 3} \\ \hline & 8 \\ \end{array}$$

✅ Quotient: $x^2 + x - 2$
✅ Remainder: $8$

---

(iii) $p(x) = x^4 - 5x + 6$, $g(x) = 2 - x^2$

✅ Rewriting $g(x)$ in standard form

$$g(x) = -x^2 + 2$$

✅ Performing Polynomial Division

$$\begin{array}{r|rrrr} & -x^2 - 1 \\ \hline - x^2 + 2 \, & x^4 - 5x + 6 \\ & \underline{x^4 - 2x^2} \\ \hline & 2x^2 - 5x + 6 \\ & \underline{2x^2 - 4} \\ \hline & -5x + 10 \\ \end{array}$$

✅ Quotient: $-x^2 - 1$
✅ Remainder: $-5x + 10$

---

2️⃣ Check whether the first polynomial is a factor of the second polynomial by dividing.

(i) $t^2 - 3$, $2t^4 + 3t^3 - 2t^2 - 9t - 12$

✅ Performing Polynomial Division

$$\begin{array}{r|rrrr} & 2t^2 + 3t + 1 \\ \hline t^2 - 3 \, & 2t^4 + 3t^3 - 2t^2 - 9t - 12 \\ & \underline{2t^4 - 6t^2} \\ \hline & 3t^3 + 4t^2 - 9t - 12 \\ & \underline{3t^3 - 9t} \\ \hline & 4t^2 - 12 \\ & \underline{4t^2 - 12} \\ \hline & 0 \\ \end{array}$$

✅ Since the remainder is $0$, $t^2 - 3$ is a factor of $2t^4 + 3t^3 - 2t^2 - 9t - 12$.

---

3️⃣ Obtain all other zeroes of $3x^4 + 6x^3 - 2x^2 - 10x - 5$, if two of its zeroes are $\frac{5}{3}$ and $-\frac{5}{3}$.

✅ Step 1: Since two zeroes are $\frac{5}{3}$ and $-\frac{5}{3}$, the factor is

$$(x - \frac{5}{3})(x + \frac{5}{3}) = x^2 - \frac{25}{9}$$

✅ Step 2: Perform Polynomial Division
Dividing $3x^4 + 6x^3 - 2x^2 - 10x - 5$ by $x^2 - \frac{25}{9}$, we get

$$3x^2 + 6x + 5$$

✅ Step 3: Solving for the remaining zeroes
Setting $3x^2 + 6x + 5 = 0$ and solving using the quadratic formula,

$$x = \frac{-6 \pm \sqrt{36 - 60}}{6} = \frac{-6 \pm \sqrt{-24}}{6}$$

Since $\sqrt{-24}$ is imaginary, the remaining two zeroes are complex.

✅ Final Zeroes:

$$\frac{5}{3}, -\frac{5}{3}, \frac{-6 + i\sqrt{24}}{6}, \frac{-6 - i\sqrt{24}}{6}$$

---

4️⃣ Find $g(x)$ when dividing $x^3 - 3x^2 + x + 2$ gives quotient $x - 2$ and remainder $-2x + 4$.

Using the division algorithm:

$$p(x) = g(x) \times q(x) + r(x)$$ $$x^3 - 3x^2 + x + 2 = g(x) \times (x - 2) + (-2x + 4)$$

✅ Solving for $g(x)$

Rearranging,

$$g(x) = \frac{x^3 - 3x^2 + x + 2 + 2x - 4}{x - 2}$$ $$= \frac{x^3 - 3x^2 + 3x - 2}{x - 2}$$

Performing division:

$$\begin{array}{r|rrrr} & x^2 - x + 1 \\ \hline x - 2 \, & x^3 - 3x^2 + 3x - 2 \\ & \underline{x^3 - 2x^2} \\ \hline & -x^2 + 3x - 2 \\ & \underline{-x^2 + 2x} \\ \hline & x - 2 \\ & \underline{x - 2} \\ \hline & 0 \\ \end{array}$$

✅ Final Answer: $g(x) = x^2 - x + 1$

---

5️⃣ Examples of polynomials satisfying the Division Algorithm

(i) $\deg p(x) = \deg q(x)$
Example:

$$p(x) = x^2 + 3x + 2, \quad g(x) = x + 1, \quad q(x) = x + 2, \quad r(x) = 0$$

(ii) $\deg q(x) = \deg r(x)$
Example:

$$p(x) = x^3 - x, \quad g(x) = x - 1, \quad q(x) = x^2, \quad r(x) = x$$

(iii) $\deg r(x) = 0$ (constant remainder)
Example:

$$p(x) = x^2 + 2x + 3, \quad g(x) = x + 1, \quad q(x) = x + 1, \quad r(x) = 2$$

---

✅ Conclusion:

✔ Polynomial division follows the rule $p(x) = g(x)q(x) + r(x)$.
✔ The remainder must have a lower degree than $g(x)$.
✔ This method helps in factorization and finding unknown zeroes.