Polynomials — Notes

2.1 Introduction to Polynomials

A polynomial is an algebraic expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.

• Variable: A symbol that represents an unknown quantity (e.g., $x,y,u,z$).
• Coefficient: A numerical or constant quantity that multiplies the variable in a term.
• Degree of a Polynomial: The highest power of the variable in a polynomial.

Types of Polynomials by Degree:

Value of a Polynomial: The value obtained by replacing the variable in a polynomial by a specific real number. If $p(x)$ is a polynomial and $k$ is a real number, then $p(k)$ is the value of the polynomial at $x=k$.

Zeroes of a Polynomial: A real number $k$ is a zero of a polynomial $p(x)$ if $p(k)=0$.

• Zero of a Linear Polynomial: For $p(x)=ax+b$, the zero is $x=-\frac{b}{a}$.

Remember

Expressions with variables in the denominator or negative exponents are not polynomials (e.g., $\frac{1}{x}$, $x^{-2}$).

Exam-smart

Identifying polynomials, their degrees, and calculating their value at a given point are fundamental. Zeroes are key for factorization and graphical interpretation.

Example 1: Evaluating a Polynomial

Given: Polynomial $p(x)=x^2-3x-4$ To find: The value of $p(x)$ at $x=2$, $x=0$, $x=-1$, and $x=4$.

Solution:

• At $x=2$:

Answer: $-6$

• At $x=0$:

Answer: $-4$

• At $x=-1$:

Answer: $0$

• At $x=4$:

Answer: $0$

Try this:

If $p(y)=3y^3-2y^2+5y-7$, find the value of $p(-2)$. Answer: $-49$

2.2 Geometrical Meaning of the Zeroes of a Polynomial

The zeroes of a polynomial $p(x)$ are the x-coordinates of the points where the graph of the equation $y=p(x)$ intersects the x-axis.

• Linear Polynomial ($y=ax+b$): The graph is a straight line. It intersects the x-axis at exactly one point, so a linear polynomial has exactly one zero. The zero is the x-intercept: $x=-b\mathrm{/}a$.

• Quadratic Polynomial ($y=a{x}^2+bx+c$): The graph is a parabola.

• If the parabola intersects the x-axis at two distinct points, there are two distinct zeroes.

• If the parabola touches the x-axis at exactly one point (i.e., two coincident points), there is one zero (a repeated root).

• If the parabola does not intersect the x-axis at all, there are no real zeroes. A quadratic polynomial has at most two zeroes.

• Cubic Polynomial ($y=a{x}^3+b{x}^2+cx+d$): The graph can intersect the x-axis at most three times. A cubic polynomial has at most three zeroes.

Generalization: A polynomial of degree $n$ has at most $n$ zeroes.

Example 2: Counting Zeroes from Graphs

Given: Graphs of $y=p(x)$. To find: The number of zeroes for each polynomial $p(x)$.

Solution: The number of zeroes is equal to the number of points where the graph intersects the x-axis.

• (i) Intersects x-axis at 1 point $\to$ 1 zero.
• (ii) Intersects x-axis at 2 points $\to$ 2 zeroes.
• (iii) Intersects x-axis at 3 points $\to$ 3 zeroes.
• (iv) Intersects x-axis at 1 point $\to$ 1 zero.
• (v) Intersects x-axis at 1 point $\to$ 1 zero.
• (vi) Intersects x-axis at 4 points $\to$ 4 zeroes.

Remember

The graph of $y=p(x)$ cutting the x-axis means the y-value is zero, which is the definition of a zero of $p(x)$.

Exam-smart

Questions will provide graphs and ask to identify the number of zeroes. This tests the understanding of the graphical interpretation of roots.

Try this:

Look at the graph provided (imagine a parabola opening upwards, with its vertex on the x-axis). How many zeroes does the corresponding quadratic polynomial have? Answer: 1

Exercise 2.1

1. Find the number of zeroes of $p(x)$ in each of the following graphs: (i) [Graph intersects x-axis at 0 points] $\to$ 0 zeroes (ii) [Graph intersects x-axis at 1 point] $\to$ 1 zero (iii) [Graph intersects x-axis at 2 points] $\to$ 2 zeroes (iv) [Graph intersects x-axis at 2 points] $\to$ 2 zeroes (v) [Graph intersects x-axis at 1 point] $\to$ 1 zero (vi) [Graph intersects x-axis at 3 points] $\to$ 3 zeroes

2.3 Relationship between Zeroes and Coefficients of a Polynomial

This section explores the connection between the values of the zeroes and the coefficients of a polynomial.

Quadratic Polynomials

For a quadratic polynomial $p(x)=a{x}^2+bx+c$, where $a\mathrm{\ne }0$, and its zeroes are $\alpha$ and $\beta$:

• Sum of Zeroes: $\alpha +\beta =-\frac{b}{a}=-\frac{\text{Coefficient of }x}{\text{Coefficient of }{x}^2}$
• Product of Zeroes: $\alpha \beta =\frac{c}{a}=\frac{\text{Constant term}}{\text{Coefficient of }{x}^2}$

Constructing a Quadratic Polynomial from its Zeroes: If $\alpha$ and $\beta$ are the zeroes, the polynomial can be written as: $p(x)=k(x^2-(\alpha +\beta )x+\alpha \beta )$, where $k$ is any non-zero real number. Often, we take $k=1$ to find one such polynomial: $p(x)=x^2-(\alpha +\beta )x+\alpha \beta$.

Cubic Polynomials

For a cubic polynomial $p(x)=a{x}^3+b{x}^2+cx+d$, where $a\mathrm{\ne }0$, and its zeroes are $\alpha$, $\beta$, and $\gamma$:

• Sum of Zeroes: $\alpha +\beta +\gamma =-\frac{b}{a}=-\frac{\text{Coefficient of }{x}^2}{\text{Coefficient of }{x}^3}$
• Sum of the products of zeroes taken two at a time: $\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}=\frac{\text{Coefficient of }x}{\text{Coefficient of }{x}^3}$
• Product of Zeroes: $\alpha \beta \gamma =-\frac{d}{a}=-\frac{\text{Constant term}}{\text{Coefficient of }{x}^3}$

Remember

The signs for the sum of zeroes are always negative, and the product of zeroes are positive/negative depending on the degree and the sign of the constant term. For a cubic polynomial, the sum of products of zeroes taken two at a time has a positive sign.

Exam-smart

Problems will ask to find zeroes and verify these relationships, or to form a polynomial given the sum and product of its zeroes.

Example 3: Finding Zeroes and Verifying Relations (Quadratic)

Given: Quadratic polynomial $p(x)=x^2+7x+10$. To find: Zeroes and verify the relationship between zeroes and coefficients.

Solution:

1. Find Zeroes: Factorize $x^2+7x+10$. We need two numbers that multiply to 10 and add to 7. These are 2 and 5.

Setting $p(x)=0$: $(x+2)(x+5)=0$ So, $x+2=0$ or $x+5=0$. $x=-2$ or $x=-5$. The zeroes are $\alpha =-2$ and $\beta =-5$.

2. Verify Relationships:

• Sum of Zeroes:

From coefficients: $-\frac{b}{a}=-\frac{7}{1}=-7$. The sum of zeroes matches.

• Product of Zeroes:

From coefficients: $\frac{c}{a}=\frac{10}{1}=10$. The product of zeroes matches.

Answer: Zeroes are -2 and -5. The relationships are verified.

Example 4: Finding Zeroes and Verifying Relations (Quadratic)

Given: Quadratic polynomial $p(x)=x^2-3$. To find: Zeroes and verify the relationship between zeroes and coefficients.

Solution:

1. Find Zeroes: Using the identity $a^2-b^2=(a-b)(a+b)$:

Setting $p(x)=0$: $(x-\sqrt{3})(x+\sqrt{3})=0$ So, $x=\sqrt{3}$ or $x=-\sqrt{3}$. The zeroes are $\alpha =\sqrt{3}$ and $\beta =-\sqrt{3}$.

2. Verify Relationships:

• Sum of Zeroes:

From coefficients: $-\frac{b}{a}=-\frac{0}{1}=0$. The sum of zeroes matches.

• Product of Zeroes:

From coefficients: $\frac{c}{a}=\frac{-3}{1}=-3$. The product of zeroes matches.

Answer: Zeroes are $\sqrt{3}$ and $-\sqrt{3}$. The relationships are verified.

Example 5: Forming a Quadratic Polynomial

Given: Sum of zeroes = -3, Product of zeroes = 2. To find: A quadratic polynomial.

Solution: Let the zeroes be $\alpha$ and $\beta$. We are given $\alpha +\beta =-3$ and $\alpha \beta =2$. The general form of a quadratic polynomial is $k(x^2-(\alpha +\beta )x+\alpha \beta )$. Substituting the given values:

Let's choose $k=1$. So, one such quadratic polynomial is $x^2+3x+2$.

Answer: $x^2+3x+2$

Example 6: Relationship for Cubic Polynomials

Given: Cubic polynomial $p(x)=2x^3-5x^2-14x+8$. Its zeroes are $4,-2,\frac{1}{2}$. To find: Verify the relationships between zeroes and coefficients.

Solution: Here, $a=2,b=-5,c=-14,d=8$. The zeroes are $\alpha =4,\beta =-2,\gamma =\frac{1}{2}$.

1. Sum of Zeroes:

From coefficients: $-\frac{b}{a}=-\frac{-5}{2}=\frac{5}{2}$. Matches.

2. Sum of products of zeroes taken two at a time:

From coefficients: $\frac{c}{a}=\frac{-14}{2}=-7$. Matches.

3. Product of Zeroes:

From coefficients: $-\frac{d}{a}=-\frac{8}{2}=-4$. Matches.

Answer: The relationships are verified.

Example 7: Verifying Zeroes of a Cubic Polynomial

Given: Cubic polynomial $p(x)=3x^3-5x^2-11x-3$. Show that $3,-1,-\frac{1}{3}$ are its zeroes and verify the relationships.

Solution: Here, $a=3,b=-5,c=-11,d=-3$. The potential zeroes are $\alpha =3,\beta =-1,\gamma =-\frac{1}{3}$.

1. Verify Zeroes:

• $p(3)=3(3{)}^3-5(3{)}^2-11(3)-3=3(27)-5(9)-33-3=81-45-33-3=0$.
• $p(-1)=3(-1{)}^3-5(-1{)}^2-11(-1)-3=3(-1)-5(1)+11-3=-3-5+11-3=0$.
• $p(-\frac{1}{3})=3(-\frac{1}{3}{)}^3-5(-\frac{1}{3}{)}^2-11(-\frac{1}{3})-3$

$=-\frac{6}{9}+\frac{11}{3}-3=-\frac{2}{3}+\frac{11}{3}-\frac{9}{3}=\frac{-2+11-9}{3}=\frac{0}{3}=0$. So, $3,-1,-\frac{1}{3}$ are indeed the zeroes.

2. Verify Relationships:

• Sum of Zeroes:

From coefficients: $-\frac{b}{a}=-\frac{-5}{3}=\frac{5}{3}$. Matches.

• Sum of products of zeroes taken two at a time:

From coefficients: $\frac{c}{a}=\frac{-11}{3}$. Matches.

• Product of Zeroes:

From coefficients: $-\frac{d}{a}=-\frac{-3}{3}=\frac{3}{3}=1$. Matches.

Answer: $3,-1,-\frac{1}{3}$ are the zeroes and the relationships are verified.

Try this:

Find a quadratic polynomial whose zeroes have a sum of $\frac{1}{4}$ and a product of $-\frac{1}{4}$. Answer: $4x^2-x-1$

Exercise 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients: (i) $p(x)=x^2-2x-8$ Solution: Zeroes: $(x-4)(x+2)=0⟹x=4,x=-2$. Sum: $4+(-2)=2$. Coeff: $-(-2)\mathrm{/}1=2$. Product: $4\times (-2)=-8$. Coeff: $-8\mathrm{/}1=-8$.

(ii) $p(s)=4s^2-4s+1$ Solution: Zeroes: $(2s-1{)}^2=0⟹s=1\mathrm{/}2,s=1\mathrm{/}2$. Sum: $1\mathrm{/}2+1\mathrm{/}2=1$. Coeff: $-(-4)\mathrm{/}4=1$. Product: $(1\mathrm{/}2)\times (1\mathrm{/}2)=1\mathrm{/}4$. Coeff: $1\mathrm{/}4$.

(iii) $p(x)=6x^2-3-7x=6x^2-7x-3$ Solution: Zeroes: $(3x+1)(2x-3)=0⟹x=-1\mathrm{/}3,x=3\mathrm{/}2$. Sum: $-1\mathrm{/}3+3\mathrm{/}2=-2\mathrm{/}6+9\mathrm{/}6=7\mathrm{/}6$. Coeff: $-(-7)\mathrm{/}6=7\mathrm{/}6$. Product: $(-1\mathrm{/}3)\times (3\mathrm{/}2)=-3\mathrm{/}6=-1\mathrm{/}2$. Coeff: $-3\mathrm{/}6=-1\mathrm{/}2$.

(iv) $p(u)=4u^2+8u$ Solution: Zeroes: $4u(u+2)=0⟹u=0,u=-2$. Sum: $0+(-2)=-2$. Coeff: $-(8)\mathrm{/}4=-2$. Product: $0\times (-2)=0$. Coeff: $0\mathrm{/}4=0$.

(v) $p(t)=t^2-15$ Solution: Zeroes: $(t-\sqrt{15})(t+\sqrt{15})=0⟹t=\sqrt{15},t=-\sqrt{15}$. Sum: $\sqrt{15}+(-\sqrt{15})=0$. Coeff: $0\mathrm{/}1=0$. Product: $(\sqrt{15})\times (-\sqrt{15})=-15$. Coeff: $-15\mathrm{/}1=-15$.

(vi) $p(x)=3x^2-x-4$ Solution: Zeroes: $(3x-4)(x+1)=0⟹x=4\mathrm{/}3,x=-1$. Sum: $4\mathrm{/}3+(-1)=1\mathrm{/}3$. Coeff: $-(-1)\mathrm{/}3=1\mathrm{/}3$. Product: $(4\mathrm{/}3)\times (-1)=-4\mathrm{/}3$. Coeff: $-4\mathrm{/}3$.

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively: (i) Sum = $\frac{1}{4}$, Product = $-\frac{1}{4}$ Answer: $x^2-(\frac{1}{4})x+(-\frac{1}{4})=0⟹4x^2-x-1$.

(ii) Sum = $\sqrt{2}$, Product = $\frac{1}{3}$ Answer: $x^2-(\sqrt{2})x+\frac{1}{3}=0⟹3x^2-3\sqrt{2}x+1$.

(iii) Sum = 0, Product = 5 Answer: $x^2-(0)x+5=0⟹x^2+5$.

(iv) Sum = 1, Product = 1 Answer: $x^2-(1)x+1=0⟹x^2-x+1$.

(v) Sum = $-\frac{1}{4}$, Product = $\frac{1}{4}$ Answer: $x^2-(-\frac{1}{4})x+\frac{1}{4}=0⟹4x^2+x+1$.

(vi) Sum = 4, Product = 1 Answer: $x^2-(4)x+1=0⟹x^2-4x+1$.

2.4 Summary

• A polynomial's degree is the highest power of its variable.
• Polynomials of degree 1, 2, and 3 are called linear, quadratic, and cubic, respectively.
• A polynomial of degree $n$ has at most $n$ zeroes.
• The zeroes of a polynomial $p(x)$ are the x-coordinates where the graph of $y=p(x)$ crosses the x-axis.
• For a quadratic polynomial $a{x}^2+bx+c$, the sum of zeroes $\alpha +\beta =-\frac{b}{a}$ and the product of zeroes $\alpha \beta =\frac{c}{a}$.
• For a cubic polynomial $a{x}^3+b{x}^2+cx+d$, the sum of zeroes $\alpha +\beta +\gamma =-\frac{b}{a}$, the sum of pairwise products $\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$, and the product of zeroes $\alpha \beta \gamma =-\frac{d}{a}$.