# CBSE 10th Mathematics | Introduction of Polynomials

## POLYNOMIALS

An expression $\displaystyle p(x)$ of the form $\displaystyle p(x)={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+...+{{a}_{n}}$ where all$\displaystyle {{a}_{0}},\,\,{{a}_{1}},\,\,{{a}_{2}},\,....,\,\,{{a}_{n}}$ are real numbers and n is a non-negative integer, is called a polynomial.

The degree of a polynomial in one variable is the greatest exponent of that variable.

$\displaystyle {{a}_{0}},\,\,{{a}_{1}},\,\,{{a}_{2}}\,...,\,\,{{a}_{n}}$ are called the co-efficients of the polynomial $\displaystyle p(x)$.

an is called constant term.

## DEGREE OF A POLYNOMIAL

The exponent of the term with the highest power in a polynomial is known as its degree.

$\displaystyle f(x)=8{{x}^{3}}-2{{x}^{2}}+8x-21$ and $\displaystyle g(x)=9{{x}^{2}}-3x+12$ are polynomials of degree 3 and 2 respectively.

Thus, $\displaystyle f(x)={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+\ldots +{{a}_{n-1}}x+{{a}_{n}}$ is a polynomial of degree n, if $\displaystyle {{a}_{0}}\ne 0.$

On the basis of degree of a polynomial, we have following standard names for the polynomials.

A polynomial of degree 1 is called a linear polynomial. Example: $\displaystyle 2x+3,\,\,\frac{1}{3}u+7$ etc.

A polynomial of degree 2 is called a quadratic polynomial. Example: $\displaystyle {{x}^{2}}+2x+3$, $\displaystyle {{y}^{2}}-9$etc.

A polynomial of degree 3 is called a cubic polynomial. Example: $\displaystyle {{x}^{3}}+7x-3$, $\displaystyle -{{x}^{3}}+{{x}^{2}}+\sqrt{3}x$ etc.

A polynomial of degree 4 is called a biquadratic polynomial. Example: $\displaystyle 3{{u}^{4}}-5{{u}^{3}}+2{{u}^{2}}+7$.

## VALUE OF A POLYNOMIAL

If $\displaystyle f(x)$ is a polynomial and a is any real number, then the real number obtained by replacing x by a in $\displaystyle f(x)$ is called the value of $\displaystyle f(x)$ at x = a and is denoted by $\displaystyle f(\alpha )$.

e.g. : Value of $\displaystyle p(x)=5{{x}^{2}}-3x+7$ at $\displaystyle x=1$ will be

\    $\displaystyle p(1)=5{{(1)}^{2}}-3(1)+7$

$\displaystyle =5-3+7=9$

## ZEROS OF A POLYNOMIAL

A real number a is a zero of polynomial $\displaystyle f(x)$ if $\displaystyle f(\alpha )$ = 0.

The zero of a linear polynomial $\displaystyle ax+b$ is $\displaystyle -\frac{b}{a}$. i.e. $\displaystyle -\frac{Constant\,\text{term}}{\text{Coefficient of }x}$

Geometrically zero of a polynomial is the point where the graph of the function cuts or touches x-axis.

When the graph of the polynomial does not meet the x-axis at all, the polynomial has no real zero.

## SIGNS OF COEFFICIENTS OF A QUADRATIC POLYNOMIAL

The graphs of $\displaystyle y=a{{x}^{2}}+bx+c$ are given in figure. Identify the signs of a, b and c in each of the following:

(i)    We observe that $\displaystyle y=a{{x}^{2}}+bx+c$ represents a parabola opening downwards. Therefore, a < 0. We observe that the turning point $\displaystyle \left( -\frac{b}{2a},\,-\frac{D}{4a} \right)$ of the parabola is in first quadrant where $\displaystyle D={{b}^{2}}-4ac$

\    $\displaystyle -\frac{b}{2a}>0$ Þ $\displaystyle -b<0$ Þ $\displaystyle b>0$                     $\displaystyle [\because \,a<0]$

Parabola $\displaystyle y=a{{x}^{2}}+bx+c$ cuts y-axis at Q. On y-axis, we have x = 0.        Putting x = 0 in $\displaystyle y=a{{x}^{2}}+bx+c,$ we get $\displaystyle y=c.$

So, the coordinates of Q are (0, c). As Q lies on the positive direction of y-axis. Therefore, c > 0.

Hence,        $\displaystyle a<0,\,\,b>0$ and $\displaystyle c>0.$

(ii)    We find that $\displaystyle y=a{{x}^{2}}+bx+c$ represents a parabola opening upwards. Therefore, $\displaystyle a>0.$ The turning point of the parabola is in fourth quadrant.

\    $\displaystyle \frac{-b}{2a}>0\Rightarrow -b>0\Rightarrow b<0.$     $\displaystyle [\because \,a>0]$

Parabola $\displaystyle y=a{{x}^{2}}+bx+c$ cuts y-axis at Q and y-axis. We have x = 0. Therefore, on putting x = 0 in $\displaystyle y=a{{x}^{2}}+bx+c,$ we get $\displaystyle y=c.$

So, the coordinates of Q are (0, c). As Q lies on negative y-axis. Therefore, c < 0.

Hence, a > 0, b < 0 and c < 0.

(iii)    Clearly, $\displaystyle y=a{{x}^{2}}+bx+c$ represents a parabola opening upwards.

Therefore, a > 0. The turning point of the parabola lies on OX.

\    $\displaystyle -\frac{b}{2a}>0\Rightarrow -b>0\Rightarrow b<0$     $\displaystyle [\because \,a>0]$

The parabola $\displaystyle y=a{{x}^{2}}+bx+c$ cuts y-axis at Q which lies on positive y-axis. Putting
x = 0 in $\displaystyle y=a{{x}^{2}}+bx+c,$ we get y = c. So, the coordinates of Q are (0, c). Clearly, Q lies on OY.

\        c > 0.

Hence,        a > 0, b < 0, and c > 0.

(iv)    The parabola $\displaystyle y=a{{x}^{2}}+bx+c$ opens downwards. Therefore, a < 0.

The turning point $\displaystyle \left( -\frac{b}{2a},-\frac{D}{4a} \right)$ of the parabola is on negative x-axis,

\    $\displaystyle -\frac{b}{2a}<0\,\,\Rightarrow \,\,\,b<0$        $\displaystyle [\because \,a<0]$

Parabola $\displaystyle y=a{{x}^{2}}+bx+c$ cuts y-axis at Q (0, c) which lies on negative y-axis. Therefore, c < 0.

Hence,     $\displaystyle a<0,\,\,b<0$ and $\displaystyle c<0.$

(v) We notice that the parabola $\displaystyle y=a{{x}^{2}}+bx+c$ opens upwards. Therefore, a > 0.

Turning point $\displaystyle \left( -\frac{b}{2a},-\frac{D}{4a} \right)$ of the parabola lies in the first quadrant.

\    $\displaystyle -\frac{b}{2a}>0\,\,\,\Rightarrow \,\,\,\frac{b}{2a}<0\,\,\,\,\Rightarrow b<0$ $\displaystyle [\because \,a>0]$

As Q (0, c) lies on positive y-axis. Therefore, c > 0.

Hence, $\displaystyle a>0,\,\,b<0$ and c > 0.

(vi)    Clearly, $\displaystyle a<0\,\,$

$\displaystyle \,Q\left( -\frac{b}{2a},-\frac{D}{4a} \right)$ lies in the fourth quadrant.

\ $\displaystyle -\frac{b}{2a}>0\,\,\,\Rightarrow \,\,\,\frac{b}{2a}<0\,\,\,\Rightarrow \,\,\,b>0$    $\displaystyle [\because \,a<0]$

As Q (0, c) lies on negative y-axis. Therefore,
c < 0.

Hence, $\displaystyle a<0,\,\,b>0$ and c < 0.

### Question:

The graphs of $\displaystyle y=p(x)$ are given below for some polynomial $\displaystyle p(x)$. Find the number of zeros of $\displaystyle p(x)$ in each case.

### Solution:

(i)    The polynomial represented in (i) has no zero because its graph does not intersect x-axis at any point.

(ii)    The polynomial represented in (ii) has one zero because its graph intersects x-axis at one point.

(iii)    The polynomial represented in (iii) has three zeros because its graph intersects x-axis at three points.

(iv)    The polynomial represented in (iv) has two zeros because its graph intersect x-axis at two points.

(v)    The polynomial represented in (v) has four zeros because its graph intersect x-axis at four points.

(vi)    The polynomial represented in (vi) has three zeros because its graph intersects x-axis at three points.

# CBSE 10th Mathematics | Decimal Representation Of Rational Numbers

## Decimal Representation Of Rational Numbers

Theorem:

Let $\displaystyle x=\frac{p}{q}$ be a rational number such that $\displaystyle q\ne 0$ and prime factorization of q is of the form $\displaystyle {{2}^{n}}\times {{5}^{m}}$ where m, n are non-negative integers then x has a decimal representation which terminates.

For example :     $\displaystyle 0.275=\frac{275}{{{10}^{3}}}=\frac{{{5}^{2}}\times 11}{{{2}^{3}}\times {{5}^{3}}}=\frac{11}{{{2}^{3}}\times 5}=\frac{11}{40}$

Theorem:

Let $\displaystyle x=\frac{p}{q}$ be a rational number such that $\displaystyle q\ne 0$ and prime factorization of q is not of the form $\displaystyle {{2}^{m}}\times {{5}^{n}}$, where m, n are non-negative integers, then x has a decimal expansion which is non-terminating repeating.

For example :     $\displaystyle \frac{5}{3}=1.66666...$

## Solved Examples Based on Decimal Representation Of Rational Numbers

### Question:

Without actually calculating, state whether the following rational numbers have a terminating or non-terminating repeating decimal expansion.

(i)     $\displaystyle \frac{27}{343}$            (ii)    $\displaystyle \frac{19}{1600}$        (iii)    $\displaystyle \frac{129}{{{2}^{2}}\times {{5}^{5}}\times {{3}^{2}}}$

Hint: If the denominator is of the form $\displaystyle {{2}^{m}}\times {{5}^{n}}$ for some non negative integer m and n, then rational number has terminating decimal otherwise non terminating.

### Solution:

(i)    $\displaystyle \frac{27}{343}=\frac{27}{{{7}^{3}}}$

Since $\displaystyle q={{7}^{3}}$ which is not of the form $\displaystyle {{2}^{m}}\times {{5}^{n}}$.

\    It has non terminating decimal representation.

(ii)    $\displaystyle \frac{19}{1600}=\frac{19}{{{2}^{6}}\times {{5}^{2}}}$

Since q = $\displaystyle {{2}^{6}}\times {{5}^{2}}$ which is of the form $\displaystyle {{2}^{m}}\times {{5}^{n}}$.

\    It has a terminating decimal representation.

(iii)    $\displaystyle \frac{129}{{{2}^{2}}\times {{5}^{5}}\times {{3}^{2}}}$

Since $\displaystyle q={{2}^{2}}\times {{5}^{5}}\times {{3}^{2}}$ is not of the form $\displaystyle {{2}^{m}}\times {{5}^{n}}$. It has a non-terminating decimal representation.

### Question:

What can you say about the prime factorization of the denominators of the following rationales:

(i)     36.12345            (ii)    $\displaystyle 36.\overline{5678}$

### Solution:

(i)    Since 36.12345 has terminating decimal expansion. So, its denominator is of the form $\displaystyle {{2}^{m}}\times {{5}^{n}}$ where m, n are non-negative integers.

(ii)    Since $\displaystyle 36.\overline{5678}$has non terminating repeating decimal expansion. So, its denominator has factors other than 2 or 5.