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}}-9etc.

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.