Class 10 Mathematics Chapter 2 Notes: Polynomials

Contents hide

1 Polynomials and Their Degree

2 Value of a Polynomial and Its Zeroes

2.1 The zero of a linear polynomial

3 Geometrical Meaning of the Zeroes

3.1 Linear polynomial: a straight line, exactly one zero

3.2 Quadratic polynomial: a parabola

3.3 Cubic polynomial: at most three zeroes

3.4 Reading the number of zeroes from a graph

4 Relationship between Zeroes and Coefficients

4.1 Two quadratics worked out

4.2 The general result for a quadratic

4.3 Building a quadratic from the sum and product of its zeroes

4.4 Cubic polynomials: three relationships

What you will learn in this chapter

Name a polynomial as linear, quadratic or cubic just by looking at its degree
Find the value of a polynomial at a number, and check whether that number is a zero
Read the number of zeroes straight off a graph
Find the zeroes of a quadratic and verify the sum and product formulas
Build a quadratic polynomial when the sum and product of its zeroes are given

degreelinearquadraticcubiczero of a polynomialparabolacoefficientsum and product of zeroes

1 Polynomials and Their Degree

You met polynomials in one variable in Class IX. A variable is a letter, such as $x$ , that stands for any number. A polynomial is built only from a variable raised to whole-number powers ( $x^0, x^1, x^2, \dots$ ), multiplied by numbers and added together. We name a polynomial in $x$ as $p(x)$ , read “p of x”: the $p$ is just its name, and the $(x)$ tells you the variable. The words below are used all through this chapter.

Word	Meaning	In $p(x) = 5x^3 – 4x^2 + x – 7$
Term	Each part joined by $+$ or $–$	$5x^3$ , $-4x^2$ , $x$ , $-7$
Coefficient	The number multiplying a power of $x$	coefficient of $x^3$ is 5, of $x^2$ is $-4$ , of $x$ is 1
Constant term	The term with no $x$ in it	$-7$

The constant term is also counted as a coefficient (it is the coefficient of $x^0 = 1$ ). The most important thing about a polynomial is its degree.

Learn by heartDefinition 1

If $p(x)$ is a polynomial in $x$ , the highest power of $x$ in $p(x)$ is called the degree of the polynomial $p(x)$ .

Expression	Variable	Degree
$4x + 2$	$x$	1
$2y^2 – 3y + 4$	$y$	2
$5x^3 – 4x^2 + x – \sqrt{2}$	$x$	3
$7u^6 – \frac{3}{2}u^4 + 4u^2 + u – 8$	$u$	6

Common Mistake

$\dfrac{1}{x-1}$ , $\sqrt{x} + 2$ and $\dfrac{1}{x^2+2x+3}$ are not polynomials. $\sqrt{x}$ is $x$ to the power $\frac{1}{2}$ , and a variable in the denominator gives a negative power (for example $\frac{1}{x} = x^{-1}$ ). Neither is a whole-number power, so none of the three is a polynomial.

Polynomials of degree 1, 2 and 3 get their own names, and these three types are the focus of this chapter. In each general form below, $a$ , $b$ , $c$ , $d$ stand for fixed real numbers (any numbers on the number line, fractions and roots like $\sqrt{2}$ included), while $x$ is the variable.

Learn by heartDefinition 2

A polynomial of degree 1 is called a linear polynomial. Its general form is $ax + b$ , where $a$ and $b$ are real numbers and $a \ne 0$ .

Learn by heartDefinition 3

A polynomial of degree 2 is called a quadratic polynomial. Any quadratic polynomial in $x$ is of the form $ax^2 + bx + c$ , where $a$ , $b$ , $c$ are real numbers and $a \ne 0$ .

Learn by heartDefinition 4

A polynomial of degree 3 is called a cubic polynomial. Its general form is $ax^3 + bx^2 + cx + d$ , where $a$ , $b$ , $c$ , $d$ are real numbers and $a \ne 0$ .

Type	Degree	General form	Examples (from your NCERT textbook)
Linear	1	$ax + b$	$2x – 3$ , $\sqrt{3}x + 5$ , $y + \sqrt{2}$ , $x – \frac{2}{11}$ , $3z + 4$ , $\frac{2}{3}u + 1$
Quadratic	2	$ax^2 + bx + c$	$2x^2 + 3x – \frac{2}{5}$ , $y^2 – 2$ , $2 – x^2 + \sqrt{3}x$ , $\frac{u}{3} – 2u^2 + 5$ , $\sqrt{5}v^2 – \frac{2}{3}v$ , $4z^2 + \frac{1}{7}$
Cubic	3	$ax^3 + bx^2 + cx + d$	$2 – x^3$ , $x^3$ , $\sqrt{2}x^3$ , $3 – x^2 + x^3$ , $3x^3 – 2x^2 + x – 1$

Did you know?

The name “quadratic” comes from the word “quadrate”, which means “square”. A quadratic polynomial is one whose highest power is a square, $x^2$ .

Exam Tip

Always write the condition $a \ne 0$ with a general form. If $a = 0$ , the term with the highest power disappears and the degree changes, so the polynomial is no longer of that type. Careful: $2x + 5 – x^2$ and $x^3 + 1$ look short and simple, but they are not linear, because their degrees are 2 and 3.

2 Value of a Polynomial and Its Zeroes

Learn by heartDefinition 5

If $p(x)$ is a polynomial in $x$ and $k$ is any real number, the value obtained by replacing $x$ by $k$ in $p(x)$ is called the value of $p(x)$ at $x = k$ , and is written $p(k)$ .

In simple words: put $k$ in place of $x$ and work the answer out.

Given

$p(x) = x^2 – 3x – 4$

To find

$p(2)$ , $p(0)$ , $p(-1)$ , $p(4)$

Solved Example

$p(2) = 2^2 – 3 \times 2 – 4 = 4 – 6 – 4 = -6$
$p(0) = 0 – 0 – 4 = -4$
$p(-1) = (-1)^2 – \{3 \times (-1)\} – 4 = 1 + 3 – 4 = 0$
$p(4) = 4^2 – (3 \times 4) – 4 = 16 – 12 – 4 = 0$

At $x = -1$ and $x = 4$ the polynomial becomes zero. Such numbers have a special name.

Learn by heartDefinition 6

A real number $k$ is said to be a zero of a polynomial $p(x)$ if $p(k) = 0$ . So $-1$ and $4$ are the zeroes of $x^2 – 3x – 4$ .

2.1 The zero of a linear polynomial

For $p(x) = 2x + 3$ : if $k$ is a zero, then $2k + 3 = 0$ , so $k = -\frac{3}{2}$ . The same steps work for any linear polynomial.

p(k) = ak + b = 0

k

is a zero, so the value is 0

ak = -b

move

b

to the other side

k = -\dfrac{b}{a}

divide by

a

(allowed, since

a \ne 0

)

Formula

\text{Zero of } ax + b = \dfrac{-b}{a} = \dfrac{-(\text{Constant term})}{\text{Coefficient of } x}

So the zero of a linear polynomial is decided entirely by its coefficients.

Common Mistake

Writing the zero of $ax + b$ as $\frac{b}{a}$ . The minus sign is part of the formula: the zero of $2x + 3$ is $-\frac{3}{2}$ , not $\frac{3}{2}$ . Check by substituting: $2\left(-\frac{3}{2}\right) + 3 = 0$ .

Are the zeroes of quadratic and cubic polynomials also linked to their coefficients? Yes, and Section 4 shows how. First, Section 3 shows what a zero looks like on a graph.

Syllabus note

NCERT’s introduction still says “We will also study the division algorithm for polynomials”. This reprint of the book has no section and no exercise on it (dividing one polynomial by another), so it is not covered in these notes.

3 Geometrical Meaning of the Zeroes

To draw the graph of $y = p(x)$ , we plot the point $(x, p(x))$ for many values of $x$ . On every graph, O is the origin, the point $(0, 0)$ where the two axes cross, and the x-coordinate of a point is its first number. The key idea: the zeroes of $p(x)$ are exactly the x-coordinates of the points where the graph of $y = p(x)$ meets the x-axis. We check this for linear, quadratic and cubic polynomials in turn.

3.1 Linear polynomial: a straight line, exactly one zero

The graph of $y = ax + b$ is a straight line. For $y = 2x + 3$ two points are enough: $x = -2$ gives $y = -1$ , and $x = 2$ gives $y = 7$ .

Graph 1 (NCERT Fig. 2.1): the line y = 2x + 3 passes through (−2, −1) and (2, 7). It meets the x-axis mid-way between x = −1 and x = −2, at (−3/2, 0). And −3/2 is exactly the zero of 2x + 3. — **Graph 1 (NCERT Fig. 2.1):** the line $y = 2x + 3$ passes through $(-2, -1)$ and $(2, 7)$ . It meets the x-axis mid-way between $x = -1$ and $x = -2$ , at $\left(-\frac{3}{2}, 0\right)$ . And $-\frac{3}{2}$ is exactly the zero of $2x + 3$ .

In general, the line $y = ax + b$ ( $a \ne 0$ ) meets the x-axis at exactly one point, $\left(-\frac{b}{a}, 0\right)$ . So a linear polynomial has exactly one zero.

3.2 Quadratic polynomial: a parabola

Take $y = x^2 – 3x – 4$ and work out $y$ for a few values of $x$ (NCERT Table 2.1):

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$
$y = x^2 – 3x – 4$	6	0	$-4$	$-6$	$-6$	$-4$	0	6

Plotting these points gives a U-shaped curve. For any quadratic $ax^2 + bx + c$ , the graph opens upwards (a U shape, like Graph 2 below) if $a > 0$ and downwards (an upside-down U) if $a \lt 0$ .

Learn by heartDefinition 7

The U-shaped (or upside-down U) graph of $y = ax^2 + bx + c$ ( $a \ne 0$ ) is called a parabola. It opens upwards when $a > 0$ and downwards when $a \lt 0$ .

The table shows $y = 0$ at $x = -1$ and $x = 4$ , and these are exactly the points where the parabola crosses the x-axis. This is true for every quadratic: its zeroes are precisely the x-coordinates of the points where its parabola meets the x-axis. A parabola can meet the x-axis in only three ways.

Cuts at two points

The graph cuts the x-axis at two different (distinct) points, named A and A′. Their x-coordinates are the two zeroes.

Touches at one point

A and A′ come together and sit on top of each other (NCERT calls them “coincident” points), so the graph just touches the x-axis at one point A. The two zeroes are equal: in NCERT’s words, “two equal zeroes (i.e., one zero)”.

iii

Never meets it

The graph is completely above or completely below the x-axis. There is no zero.

Graph 2 (NCERT Fig. 2.2), Case (i): y = x² − 3x − 4 cuts the x-axis at A(−1, 0) and A′(4, 0), so its two zeroes are −1 and 4 — **Graph 2 (NCERT Fig. 2.2), Case (i):** $y = x^2 – 3x – 4$ cuts the x-axis at A $(-1, 0)$ and A′ $(4, 0)$ , so its two zeroes are $-1$ and $4$

Graph 3, Case (ii): y = (x − 2)² just touches the x-axis at A(2, 0), so its only zero is 2 (an extra example, not in NCERT) — **Graph 3, Case (ii):** $y = (x – 2)^2$ just touches the x-axis at A $(2, 0)$ , so its only zero is $2$ (an extra example, not in NCERT)

Graph 4, Case (iii): y = x² + 2 (completely above the x-axis) and y = −x² − 1 (completely below it) never meet the x-axis, so neither has a zero (extra examples, not in NCERT) — **Graph 4, Case (iii):** $y = x^2 + 2$ (completely above the x-axis) and $y = -x^2 – 1$ (completely below it) never meet the x-axis, so neither has a zero (extra examples, not in NCERT)

Memory Trick

Cuts: 2 · Touches: 1 · Misses: 0

These are the only three possibilities, so a polynomial of degree 2 has at most two zeroes.

Exam Tip

NCERT’s own footnote: plotting graphs of quadratic or cubic polynomials is not meant to be done by students and is not evaluated. What you do need is to read a given graph: count where it meets the x-axis.

3.3 Cubic polynomial: at most three zeroes

For $y = x^3 – 4x$ (NCERT Table 2.2):

$x$	$-2$	$-1$	$0$	$1$	$2$
$y = x^3 – 4x$	0	3	0	$-3$	0

Graph 5 (NCERT Fig. 2.6): y = x³ − 4x meets the x-axis only at (−2, 0), (0, 0) and (2, 0), so its zeroes are −2, 0 and 2 — **Graph 5 (NCERT Fig. 2.6):** $y = x^3 – 4x$ meets the x-axis only at $(-2, 0)$ , $(0, 0)$ and $(2, 0)$ , so its zeroes are $-2$ , $0$ and $2$

Graph 6 (NCERT Figs. 2.7 and 2.8): y = x³ meets the x-axis only at the origin (one zero, 0), while y = x³ − x² = x²(x − 1) touches the x-axis at x = 0 and crosses it at x = 1, so it has two different zeroes, 0 and 1 — **Graph 6 (NCERT Figs. 2.7 and 2.8):** $y = x^3$ meets the x-axis only at the origin (one zero, $0$ ), while $y = x^3 – x^2 = x^2(x – 1)$ touches the x-axis at $x = 0$ and crosses it at $x = 1$ , so it has two different zeroes, 0 and 1

Cubic polynomial	Zeroes	Number of zeroes
$x^3 – 4x$	$-2,\ 0,\ 2$	3
$x^3$	$0$	1
$x^3 – x^2 = x^2(x – 1)$	$0,\ 1$	2

So a cubic can have 1, 2 or 3 zeroes, but never more: any polynomial of degree 3 has at most three zeroes. The pattern extends to every degree.

In general, given a polynomial $p(x)$ of degree $n$ , the graph of $y = p(x)$ intersects the x-axis at atmost $n$ points. Therefore, a polynomial $p(x)$ of degree $n$ has at most $n$ zeroes.

Remark in the NCERT textbook, Section 2.2

3.4 Reading the number of zeroes from a graph

NCERT practises this skill in its Example 1 (Fig. 2.9) and in Exercise 2.1 (Fig. 2.10); each figure has six graphs. Count the points where the curve meets the x-axis, whether it crosses there or only touches.

Graph 7: six extra graphs of y = p(x) for counting practice (not from NCERT); (a) y = 2 is a constant polynomial (2 = 2x⁰, degree 0), and (e), (f) have degrees 4 and 5; the red dots mark every point where the curve meets the x-axis, giving 0, 1, 3, 2, 4 and 3 zeroes — **Graph 7:** six extra graphs of $y = p(x)$ for counting practice (not from NCERT); (a) $y = 2$ is a constant polynomial ( $2 = 2x^0$ , degree 0), and (e), (f) have degrees 4 and 5; the red dots mark every point where the curve meets the x-axis, giving 0, 1, 3, 2, 4 and 3 zeroes

✓ Count these

Every point where the curve crosses the x-axis
A point where the curve only touches the x-axis and turns back, counted once (like (f) above)

✗ Never count these

The point where the curve crosses the y-axis
A horizontal line above or below the x-axis: it has no zero at all (like (a) above)

Exam Tip

A question can show a graph and ask “the number of zeroes of $p(x)$ is”. In NCERT’s Example 1 the six graphs have 1, 2, 3, 1, 1 and 4 zeroes; a graph with 4 zeroes must belong to a polynomial of degree at least 4.

4 Relationship between Zeroes and Coefficients

We saw that the zero of $ax + b$ is $-\frac{b}{a}$ . Now we find a similar link for quadratics. First, two examples solved by splitting the middle term (the Class IX method): for $ax^2 + bx + c$ , break $bx$ into two terms whose product is $ax^2 \times c$ and whose sum is $bx$ , then take out common factors.

4.1 Two quadratics worked out

Solved Example

Q. Find the zeroes of $p(x) = 2x^2 – 8x + 6$ and compare them with the coefficients.

Step 1: Split $-8x$ into two terms whose product is $6 \times 2x^2 = 12x^2$ and whose sum is $-8x$ : these are $-6x$ and $-2x$ .
Step 2: $2x^2 – 6x – 2x + 6 = 2x(x – 3) – 2(x – 3) = (2x – 2)(x – 3) = 2(x – 1)(x – 3)$
Step 3: $p(x) = 0$ when $x = 1$ or $x = 3$ . Zeroes: $1$ and $3$ .
Compare with the coefficients ( $a = 2$ , $b = -8$ , $c = 6$ ): sum $= 1 + 3 = 4 = \dfrac{-(-8)}{2}$ ; product $= 1 \times 3 = 3 = \dfrac{6}{2}$ .

Solved Example

Q. Find the zeroes of $p(x) = 3x^2 + 5x – 2$ and compare them with the coefficients.

$3x^2 + 6x – x – 2 = 3x(x + 2) – 1(x + 2) = (3x – 1)(x + 2)$ . Zeroes: $\frac{1}{3}$ and $-2$ .
Compare: sum $= \frac{1}{3} + (-2) = -\frac{5}{3} = \dfrac{-(5)}{3}$ ; product $= \frac{1}{3} \times (-2) = -\frac{2}{3} = \dfrac{-2}{3}$ .

In both, the sum of zeroes is $\frac{-(\text{coefficient of } x)}{\text{coefficient of } x^2}$ and the product is $\frac{\text{constant term}}{\text{coefficient of } x^2}$ . Here is why this is always true.

4.2 The general result for a quadratic

We name the zeroes with Greek letters: $\alpha$ (alpha) and $\beta$ (beta). If $\alpha$ and $\beta$ are the zeroes of $ax^2 + bx + c$ , then $(x – \alpha)$ and $(x – \beta)$ are its factors (expressions that multiply together to give it). You saw this in the first example above: zeroes 1 and 3 came from the factors $(x – 1)$ and $(x – 3)$ .

ax^2 + bx + c = k(x – \alpha)(x – \beta)

k

is a constant, a fixed number

= k[x^2 – (\alpha + \beta)x + \alpha\beta]

multiply out the brackets

= kx^2 – k(\alpha + \beta)x + k\alpha\beta

multiply by

k

a = k,\quad b = -k(\alpha + \beta),\quad c = k\alpha\beta

both sides are the same polynomial, so the numbers in front of

x^2

, of

x

, and the constant terms must match

\alpha + \beta = \dfrac{-b}{a},\quad \alpha\beta = \dfrac{c}{a}

put

k = a

and divide

Formula

\alpha + \beta = \dfrac{-b}{a} = \dfrac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2} \qquad \alpha\beta = \dfrac{c}{a} = \dfrac{\text{Constant term}}{\text{Coefficient of } x^2}

\alpha

\beta

are the zeroes of the quadratic polynomial

ax^2 + bx + c

a \ne 0

What the symbols mean

\alpha

alpha: the first zero1st

\beta

beta: the second zero2nd

\gamma

gamma: the third zero, used for cubics3rd

NCERT Example 2

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

$x^2 + 7x + 10 = (x + 2)(x + 5)$ , so the zeroes are $-2$ and $-5$ .
Sum $= -2 + (-5) = -7 = \dfrac{-(7)}{1}$ ✓ Product $= (-2) \times (-5) = 10 = \dfrac{10}{1}$ ✓

NCERT Example 3

Q. Find the zeroes of $x^2 – 3$ and verify the relationship.

Using $A^2 – B^2 = (A – B)(A + B)$ : $x^2 – 3 = (x – \sqrt{3})(x + \sqrt{3})$ , so the zeroes are $\sqrt{3}$ and $-\sqrt{3}$ .
Sum $= \sqrt{3} – \sqrt{3} = 0$ , and there is no $x$ term, so $b = 0$ and $\dfrac{-b}{a} = 0$ ✓ Product $= (\sqrt{3})(-\sqrt{3}) = -3 = \dfrac{-3}{1}$ ✓

4.3 Building a quadratic from the sum and product of its zeroes

Dividing $ax^2 + bx + c$ by $a$ gives $x^2 + \frac{b}{a}x + \frac{c}{a} = x^2 – (\alpha + \beta)x + \alpha\beta$ . So one quadratic with given sum $S$ and product $P$ of zeroes is $x^2 – Sx + P$ .

NCERT Example 4

Q. Find a quadratic polynomial, the sum and product of whose zeroes are $-3$ and $2$ , respectively.

Let the polynomial be $ax^2 + bx + c$ with zeroes $\alpha$ , $\beta$ . Then $\alpha + \beta = -3 = \frac{-b}{a}$ and $\alpha\beta = 2 = \frac{c}{a}$ .
Taking $a = 1$ : $b = 3$ and $c = 2$ . So one such polynomial is $x^2 + 3x + 2$ .
Any other quadratic that fits is of the form $k(x^2 + 3x + 2)$ , where $k$ is real (and $k \ne 0$ , or it would not be a quadratic).

Exercise 2.2 practises both skills: Question 1 is “find the zeroes and verify” (section 4.2) and Question 2 is “find the polynomial from the sum and product” (this section).

Common Mistake

Writing $x^2 + Sx + P$ . The middle term takes the opposite sign of the sum: sum $-3$ gives $+3x$ . When $S$ or $P$ is a fraction, multiply every term by a suitable $k$ to clear the fractions; the zeroes do not change. For example, $S = \frac{1}{4}$ , $P = -1$ gives $x^2 – \frac{1}{4}x – 1$ , and multiplying by 4 gives $4x^2 – x – 4$ .

Exam-style application (extra, not in NCERT)

Q. If $\alpha$ , $\beta$ are the zeroes of $x^2 – 5x + 6$ , find $\alpha^2 + \beta^2$ and $\frac{1}{\alpha} + \frac{1}{\beta}$ without finding the zeroes.

$\alpha + \beta = 5$ , $\alpha\beta = 6$ .
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta = 25 – 12 = 13$
$\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{5}{6}$
Answer: $\alpha^2 + \beta^2 = 13$ and $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{5}{6}$
Check: the zeroes are 2 and 3, and $4 + 9 = 13$ , $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ ✓

4.4 Cubic polynomials: three relationships

Take $p(x) = 2x^3 – 5x^2 – 14x + 8$ . You can check that $p(x) = 0$ for $x = 4$ , $-2$ and $\frac{1}{2}$ ; for example $p(4) = 128 – 80 – 56 + 8 = 0$ . A cubic has three relationships. The middle one uses the “sum of the products of the zeroes taken two at a time”: multiply the zeroes in pairs ( $\alpha\beta$ , $\beta\gamma$ , $\gamma\alpha$ ) and add the three products. Since a cubic has at most three zeroes, these are all of them.

Quantity	Working	From the coefficients
Sum of zeroes	$4 + (-2) + \frac{1}{2} = \frac{5}{2}$	$\frac{-(-5)}{2} = \frac{-(\text{coeff. of } x^2)}{\text{coeff. of } x^3}$
Sum of the products of the zeroes taken two at a time	$4(-2) + (-2)\left(\frac{1}{2}\right) + \frac{1}{2}(4) = -8 – 1 + 2 = -7$	$\frac{-14}{2} = \frac{\text{coeff. of } x}{\text{coeff. of } x^3}$
Product of zeroes	$4 \times (-2) \times \frac{1}{2} = -4$	$\frac{-8}{2} = \frac{-(\text{constant term})}{\text{coeff. of } x^3}$

Formula

\alpha + \beta + \gamma = \dfrac{-b}{a} \qquad \alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{c}{a} \qquad \alpha\beta\gamma = \dfrac{-d}{a}

\alpha

\beta

\gamma

are the zeroes of the cubic polynomial

ax^3 + bx^2 + cx + d

a \ne 0

Memory Trick

Signs go minus, plus, minus

Read the coefficients $b, c, d$ in order and divide each by $a$ : the three results take the signs $–$ , $+$ , $–$ . For a quadratic only the first two apply: $-\frac{b}{a}$ , $+\frac{c}{a}$ .

NCERT Example 5 (marked by NCERT “not from the examination point of view”)

Q. Verify that $3$ , $-1$ , $-\frac{1}{3}$ are the zeroes of $p(x) = 3x^3 – 5x^2 – 11x – 3$ , and then verify the relationship between the zeroes and the coefficients.

Here $a = 3$ , $b = -5$ , $c = -11$ , $d = -3$ .
$p(3) = 81 – 45 – 33 – 3 = 0$ ; $p(-1) = -3 – 5 + 11 – 3 = 0$ ; $p\left(-\frac{1}{3}\right) = -\frac{1}{9} – \frac{5}{9} + \frac{11}{3} – 3 = 0$ .
Sum $= 3 – 1 – \frac{1}{3} = \frac{5}{3} = \frac{-b}{a}$ ✓; sum of the products taken two at a time $= -3 + \frac{1}{3} – 1 = -\frac{11}{3} = \frac{c}{a}$ ✓; product $= 1 = \frac{-(-3)}{3} = \frac{-d}{a}$ ✓

Read it once for understanding; the three cubic formulas themselves are in NCERT’s Summary and can be asked.

Quick Revision: read this the night before the exam

Degree 1, 2, 3 = linear, quadratic, cubic; always add $a \ne 0$
Zero of $ax + b$ is $-\frac{b}{a}$ : exactly one zero, the line meets the x-axis once
Zeroes of $p(x)$ = x-coordinates of the points where $y = p(x)$ meets the x-axis
Quadratic: 2, 1 or 0 zeroes (cuts, touches, misses), at most 2
Cubic: at most 3 zeroes; degree $n$ : at most $n$ zeroes
Quadratic: $\alpha + \beta = -\frac{b}{a}$ , $\alpha\beta = \frac{c}{a}$
From sum $S$ and product $P$ : $k(x^2 – Sx + P)$ , $k \ne 0$
Cubic: $\alpha + \beta + \gamma = -\frac{b}{a}$ , $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$ , $\alpha\beta\gamma = -\frac{d}{a}$
Not examined: plotting quadratic/cubic graphs, NCERT Example 5; the division algorithm (dividing polynomials) has no section or exercise in this reprint

More for this chapter

Important Questions MCQ / Objective NCERT Solutions Short Notes