Polynomials

Class 10 . Mathematics . Mathematics . Polynomials

Chapter 2: Polynomials

Ch 2

Mathematics

Class 10

Lecture 1

Introduction

Reference video · Lecture 1Introduction

Degree and Types of Polynomials

Definition

Degree of a polynomial — The degree is the greatest exponent of the variable with a non-zero coefficient.

Polynomial types

Polynomial: Variable exponents are non-negative integers; coefficients may be real numbers.
Not polynomials: Expressions such as $\frac{1}{x}-1$, $\sqrt{x}+2$, and $\frac{1}{x^2}+2x+3$ have a variable in the denominator or a non-integer exponent.
Linear: A polynomial of degree $1$, such as $2x-3$ or $\sqrt{3}x+5$.
Quadratic: A polynomial of degree $2$, such as $x^2-2$.
Cubic: A polynomial of degree $3$, such as $2-x^3$.

General Forms by Degree

$$\text{Linear: } ax+b,\ a\ne 0;\qquad \text{Quadratic: } ax^2+bx+c,\ a\ne 0;\qquad \text{Cubic: } ax^3+bx^2+cx+d,\ a\ne 0$$

The leading coefficient, the coefficient of the highest power, must be non-zero.

Worked example

Classify $5x^3-4x^2+x-\sqrt{2}$ and $2x+5-x^2$ by degree.

The highest exponent is $3$, so this polynomial is cubic.
$$5x^3-4x^2+x-\sqrt{2}$$
The highest exponent is $2$, so this polynomial is quadratic, not linear.
$$2x+5-x^2$$

Answer: The first polynomial is cubic; the second is quadratic.

Values and Zeroes of a Polynomial

Definition

Value of a polynomial — For a polynomial $p(x)$, its value at $x=k$ is found by substituting $k$ for $x$ and is written as $p(k)$.

Definition

Zero of a polynomial — A real number $k$ is a zero of $p(x)$ if and only if $p(k)=0$.

Zero of a Linear Polynomial

$$ax+b=0\quad\Longrightarrow\quad x=-\frac{b}{a},\quad a\ne 0$$

Set the linear polynomial equal to zero and solve for its variable.

Worked example

For $p(x)=x^2-3x-4$, find $p(2)$ and determine whether $-1$ is a zero.

Substitute $x=2$ into the polynomial and simplify.
$$p(2)=2^2-3(2)-4=4-6-4=-6$$
Substitute $x=-1$; the result is zero, so $-1$ is a zero.
$$p(-1)=(-1)^2-3(-1)-4=1+3-4=0$$

Answer: $p(2)=-6$; $-1$ is a zero because $p(-1)=0$.

Geometrical Meaning and Number of Zeroes

Definition

Graphical meaning of a zero — The zeroes of $p(x)$ are the $x$-coordinates of the points where the graph $y=p(x)$ meets the $x$-axis.

Graph behaviour

Linear: The graph of $y=ax+b$, where $a\ne0$, is a straight line that meets the $x$-axis once, at $\left(-\frac{b}{a},0\right)$.
Example: The line $y=2x+3$ passes through $(-2,-1)$ and $(2,7)$ and meets the $x$-axis at $\left(-\frac{3}{2},0\right)$.
Quadratic: The graph of $y=ax^2+bx+c$ is a parabola; it opens upwards when $a>0$ and downwards when $a<0$.
Quadratic zeroes: A parabola may cut the $x$-axis twice, touch it once (a repeated zero), or not meet it; thus a quadratic has at most two real zeroes.
Cubic examples: $x^3-4x=x(x-2)(x+2)$ has zeroes $-2,0,2$; $x^3$ has the distinct zero $0$.
More cubic examples: $x^3-x^2=x^2(x-1)$ has distinct zeroes $0$ and $1$.
General rule: A polynomial of degree $n$ has at most $n$ zeroes.

The line y=2x+3 meets the x-axis at its zero, x=-32. — The line $y=2x+3$ meets the $x$-axis at its zero, $x=-\frac{3}{2}$.

Values of $y=x^2-3x-4$

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$
$y$	$6$	$0$	$-4$	$-6$	$-6$	$-4$	$0$	$6$

The quadratic graph meets the x-axis at x=-1 and x=4. — The quadratic graph meets the $x$-axis at $x=-1$ and $x=4$.

Examples of polynomial graphs; the number of zeroes is the number of distinct points where each graph meets the x-axis. — Examples of polynomial graphs; the number of zeroes is the number of distinct points where each graph meets the $x$-axis.

Counting zeroes from graphs

Graph (i): $1$ zero.
Graph (ii): $2$ zeroes.
Graph (iii): $3$ zeroes.
Graph (iv): $1$ zero.
Graph (v): $1$ zero.
Graph (vi): $4$ zeroes.

Worked example

Find the zeroes of $x^3-4x$ and state their graphical meaning.

Factor out $x$, then factor the difference of squares.
$$x^3-4x=x(x^2-4)=x(x-2)(x+2)$$
Set each factor equal to zero to find the zeroes.
$$x=0,\quad x=2,\quad x=-2$$

Answer: The zeroes are $-2,0,2$; they are the $x$-coordinates where the graph meets the $x$-axis.

⚠ Common mistake

A graph that touches the $x$-axis still has a zero there; a repeated zero counts as one distinct zero and one point of contact.

Relations Between Zeroes and Coefficients

Zero-coefficient relations

For a quadratic with zeroes $\alpha,\beta$ and a cubic with zeroes $\alpha,\beta,\gamma$, sums and products of the zeroes are related to the coefficients.

Zeroes and Coefficients

$$\begin{aligned}\text{Quadratic } ax^2+bx+c:\quad &\alpha+\beta=-\frac{b}{a},\quad \alpha\beta=\frac{c}{a}\\\text{Cubic } ax^3+bx^2+cx+d:\quad &\alpha+\beta+\gamma=-\frac{b}{a},\quad \alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a},\quad \alpha\beta\gamma=-\frac{d}{a}\end{aligned}$$

Use these relations when the polynomial coefficients and its zeroes are known.

Worked example

Verify the zero-coefficient relations for $x^2-3x-4$.

Factor to find the zeroes.
$$x^2-3x-4=(x-4)(x+1),\quad \alpha=4,\quad \beta=-1$$
Check that the sum equals $-\frac{b}{a}$.
$$\alpha+\beta=4+(-1)=3=-\frac{-3}{1}$$
Check that the product equals $\frac{c}{a}$.
$$\alpha\beta=4(-1)=-4=\frac{-4}{1}$$

Answer: Their sum is $3=-\frac{b}{a}$ and their product is $-4=\frac{c}{a}$.

Practice

Find the degree of $7u^6-\frac{3}{2}u^4+4u^2+u-8$.

Hint: Look for the greatest exponent of $u$.

Key step: The highest power is $u^6$.

Solution: The degree is $6$.

Practice

How many distinct zeroes does $x^3-x^2$ have?

Hint: Factor the polynomial.

Key step: Write it as $x^2(x-1)$.

Solution: $x^3-x^2=x^2(x-1)$, so its distinct zeroes are $0$ and $1$; there are $2$.

Practice

A polynomial graph touches the $x$-axis at one point and does not cross it. How many distinct zeroes does it have?

Hint: Touching the axis still means the graph meets it.

Key step: The $x$-coordinate of the contact point is a zero.

Solution: It has one distinct zero.

Quick recap

The degree is the greatest exponent with a non-zero coefficient; degrees $1$, $2$, and $3$ are linear, quadratic, and cubic.
A real number $k$ is a zero of $p(x)$ exactly when $p(k)=0$.
Graphically, zeroes are the $x$-coordinates where $y=p(x)$ meets the $x$-axis.
A quadratic can have two distinct real zeroes, one repeated real zero, or no real zero; a cubic has at most three zeroes.
A polynomial of degree $n$ has at most $n$ zeroes.
For a quadratic, $\alpha+\beta=-\frac{b}{a}$ and $\alpha\beta=\frac{c}{a}$.
For a cubic, the sum, pairwise-product sum, and product of the zeroes are $-\frac{b}{a}$, $\frac{c}{a}$, and $-\frac{d}{a}$.

Lecture 2

Relationship Between Zeroes And Coefficients Of A Polynomial

Reference video · Lecture 2Relationship Between Zeroes And Coefficients Of A Polynomial

Quadratic Polynomials

Definition

Quadratic polynomial — A quadratic polynomial has the form $ax^2+bx+c$, where $a\ne 0$.

Quadratic zeroes and coefficients

If the zeroes are $\alpha$ and $\beta$, their sum is the negative ratio of the coefficient of $x$ to that of $x^2$, and their product is the ratio of the constant term to that of $x^2$.

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

Key points

Factored form: $ax^2+bx+c=a(x-\alpha)(x-\beta)$.
Coefficient comparison: The coefficients determine the sum and product of the zeroes.
Reverse use: Given the sum and product, a suitable quadratic can be formed.

Quadratic zeroes

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

Use for a quadratic polynomial $ax^2+bx+c$, with zeroes $\alpha$ and $\beta$.

Find and verify quadratic zeroes

Factorise the polynomial.
$$ax^2+bx+c=a(x-r_1)(x-r_2)$$
Set each factor equal to zero to find the zeroes.
$$x-r_1=0,\qquad x-r_2=0$$
Check their sum and product against the coefficient formulas.
$$r_1+r_2=-\frac{b}{a},\qquad r_1r_2=\frac{c}{a}$$

Worked example

Find the zeroes of $x^2-3$ and verify their relationship with the coefficients.

Factorise the polynomial using the difference of squares.
$$x^2-3=(x-\sqrt{3})(x+\sqrt{3})$$
Set each factor to zero; the zeroes are $\sqrt{3}$ and $-\sqrt{3}$.
$$x-\sqrt{3}=0\Rightarrow x=\sqrt{3},\qquad x+\sqrt{3}=0\Rightarrow x=-\sqrt{3}$$
Verify the sum using $a=1$ and $b=0$.
$$\sqrt{3}+(-\sqrt{3})=0=-\frac{0}{1}$$
Verify the product using $a=1$ and $c=-3$.
$$\sqrt{3}(-\sqrt{3})=-3=\frac{-3}{1}$$

Answer: The zeroes are $\sqrt{3}$ and $-\sqrt{3}$; their sum is $0=-b/a$ and their product is $-3=c/a$.

⚠ Common mistake

Do not omit the negative sign in the sum formula: the sum is $-b/a$, while the product is $c/a$.

Cubic Polynomials

Definition

Cubic polynomial — A cubic polynomial has the form $ax^3+bx^2+cx+d$, where $a\ne 0$.

Cubic zeroes and coefficients

If the zeroes are $\alpha$, $\beta$ and $\gamma$, their sum, pairwise-product sum and product are related to the coefficients as shown.

$$\alpha+\beta+\gamma=-\frac{b}{a},\qquad \alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a},\qquad \alpha\beta\gamma=-\frac{d}{a}$$

Key points

Factored form: $ax^3+bx^2+cx+d=a(x-\alpha)(x-\beta)(x-\gamma)$.
Pairwise-product sum: Add the three products of zeroes taken two at a time: $\alpha\beta+\beta\gamma+\gamma\alpha$.
Signs: The sum and product have negative signs in their coefficient relations; the pairwise-product sum is positive.

Cubic zeroes

$$\alpha+\beta+\gamma=-\frac{b}{a},\qquad \alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a},\qquad \alpha\beta\gamma=-\frac{d}{a}$$

Use for a cubic polynomial $ax^3+bx^2+cx+d$, with zeroes $\alpha$, $\beta$ and $\gamma$.

Verify cubic zeroes and coefficients

Check each proposed zero by substitution or by factorisation.
$$p(\alpha)=p(\beta)=p(\gamma)=0$$
Calculate the sum of the three zeroes and compare it with the coefficient formula.
$$\alpha+\beta+\gamma=-\frac{b}{a}$$
Calculate the sum of pairwise products and compare it with the coefficient formula.
$$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}$$
Calculate the product of the three zeroes and compare it with the coefficient formula.
$$\alpha\beta\gamma=-\frac{d}{a}$$

Worked example

Verify that $3$, $-1$ and $-1/3$ are zeroes of $3x^3-5x^2-11x-3$, then verify the coefficient relationships.

Substitute $x=3$ to check the first proposed zero.
$$p(3)=3(3)^3-5(3)^2-11(3)-3=81-45-33-3=0$$
Substitute $x=-1$ to check the second proposed zero.
$$p(-1)=3(-1)^3-5(-1)^2-11(-1)-3=-3-5+11-3=0$$
Substitute $x=-\frac{1}{3}$ to check the third proposed zero.
$$p\left(-\frac{1}{3}\right)=-\frac{1}{9}-\frac{5}{9}+\frac{11}{3}-3=0$$
Their sum agrees with $-b/a$.
$$3+(-1)-\frac{1}{3}=\frac{5}{3}=-\frac{-5}{3}$$
Their pairwise-product sum agrees with $c/a$.
$$3(-1)+(-1)\left(-\frac{1}{3}\right)+\left(-\frac{1}{3}\right)3=-3+\frac{1}{3}-1=-\frac{11}{3}$$
Their product agrees with $-d/a$.
$$3(-1)\left(-\frac{1}{3}\right)=1=-\frac{-3}{3}$$

Answer: All three are zeroes. Their sum is $5/3=-b/a$, their pairwise-product sum is $-11/3=c/a$, and their product is $1=-d/a$.

⚠ Common mistake

For a cubic, do not confuse the pairwise-product sum with the product of all three zeroes; their coefficient formulas are different.

Practice

Find the zeroes of $x^2+7x+10$ and verify their relationship with the coefficients.

Hint: Factorise the quadratic, then compare the zeroes' sum and product with $-b/a$ and $c/a$.

Key step: Use $x^2+7x+10=(x+2)(x+5)$.

Solution: The zeroes are $-2$ and $-5$. Their sum is $-7=-7/1=-b/a$, and their product is $10=10/1=c/a$.

Practice

Find a quadratic polynomial whose zeroes have sum $-3$ and product $2$.

Hint: For a monic quadratic, use $x^2-Sx+P$, where $S$ is the sum and $P$ the product.

Key step: Substitute $S=-3$ and $P=2$ into $x^2-Sx+P$.

Solution: One such polynomial is $x^2+3x+2=(x+1)(x+2)$. Any non-zero scalar multiple also works.

Practice

A monic cubic has zeroes $1$, $2$ and $-3$. Find the polynomial.

Hint: Find the sum, pairwise-product sum and product, then use the cubic coefficient relationships.

Key step: The sum is $0$, the pairwise-product sum is $-7$, and the product is $-6$.

Solution: The polynomial is $(x-1)(x-2)(x+3)=x^3-7x+6$.

Memory hook Prepzy tip

Quadratic sign check

For $ax^2+bx+c$, the zeroes' sum has the opposite sign to $b/a$; their product has the same sign as $c/a$.

Quick recap

For $ax^2+bx+c$, the zeroes satisfy $\alpha+\beta=-b/a$ and $\alpha\beta=c/a$.
For $ax^3+bx^2+cx+d$, the zeroes satisfy $\alpha+\beta+\gamma=-b/a$.
The cubic pairwise-product sum is $\alpha\beta+\beta\gamma+\gamma\alpha=c/a$.
The product of the cubic zeroes is $\alpha\beta\gamma=-d/a$.
The zeroes of $x^2-3$ are the irrational values $\sqrt{3}$ and $-\sqrt{3}$, and they satisfy the quadratic relationships.
Use coefficient relationships both to verify zeroes and to construct polynomials.

Formula sheet

General Forms by Degree

$$\text{Linear: } ax+b,\ a\ne 0;\qquad \text{Quadratic: } ax^2+bx+c,\ a\ne 0;\qquad \text{Cubic: } ax^3+bx^2+cx+d,\ a\ne 0$$

Zero of a Linear Polynomial

$$ax+b=0\quad\Longrightarrow\quad x=-\frac{b}{a},\quad a\ne 0$$

Zeroes and Coefficients

Quadratic zeroes

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

Cubic zeroes

$$\alpha+\beta+\gamma=-\frac{b}{a},\qquad \alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a},\qquad \alpha\beta\gamma=-\frac{d}{a}$$

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