Polynomials in Mathematics – Definition, Types, Formulas & Applications

Polynomials

An algebraic expression of the form $p (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + \dots + a_{1} x^{1} + a_{0} x^{0}$ , where (i) $a_{n} \neq = 0$ (ii) $a_{0}, a_{1}, a_{2}, \dots a_{n}$ are real numbers (iii) power of x is a non negative integer, is called a polynomial.

Where $a_{n}, a_{n - 1}, a_{n - 2}$ , are coefficients of $x^{n}, x^{n - 1}$ $x^{0}$ respectively and $a_{n} x^{n}, a_{n - 1} x^{n - 1}, a_{n - 2} x^{n - 2}, \dots .. a_{0} x^{0}$ are terms of the polynomial. Here the term $a_{n} x^{n}$ is called the leading term and its coefficient $a_{n}$ is the leading coefficient. For e.g.: $p (u) = \frac{1}{2} u^{3} - 3 u^{2} + 2 u - 4$ is a polynomial in variable $u$ . $\frac{1}{2} u^{3}, - 3 u^{2}, 2 u$ and -4 are known as terms of polynomial and $\frac{1}{2}, - 3$ and 2 are their respective coefficients.

All algebraic expressions are not polynomials.

1.0Types of polynomials:

Generally we divide the polynomials in three categories.

Polynomials classified by number of distinct variables

Number of distinct variables	Name	Example
1	Univariate	$x + 9$
2	Bivariate	$x + y + 9$
3	Trivariate	$x + y + z + 9$

Generally, a polynomial in more than one variable is called a multivariate polynomial. A second major way of classifying polynomials is by their degree.

Recall that the degree of a term is the sum of the exponents on variables, and that the degree of a polynomial is the largest degree of any one term. e.g. Polynomial $= x^{2} y^{3} + x^{7} y^{3} + x^{2} y^{2}$ here, degrees of term $x^{2} y^{3}, x^{7} y^{3}, x^{2} y^{2}$ are $5, 10, 4$ respectively. Degree of given polynomial $= 10$ Polynomials classified by degree

Degree	Name	Example
$- \infty$	Zero	0
0	(non-zero) Constant	1
1	Linear	$x + 1$
2	Quadratic	$x^{2} + 1$
3	Cubic	$p (u) = \frac{1}{2} u^{3} - 3 u^{2} + 2 u - 4$

Usually, a polynomial of degree $n$ , for $n$ greater than 3 , is called a polynomial of degree $n$ , although the phrases quartic polynomial and quantic polynomial are sometimes used for degree 4 and 5 respectively. Polynomials classified by number of non-zero terms

Number of non-zero terms	Name	Example
0	Zero polynomial	0
1	Monomial	$x^{2}$
2	Binomial	$x^{2} + 1$
3	Trinomial	$x^{2} + x + 1$

If a polynomial has only one variable, then the terms are usually written either from highest degree to lowest degree ("descending powers") or from lowest degree to highest degree ("ascending powers").

2.0Value of a polynomial

If $p (x)$ is a polynomial in variable $x$ and $a$ is any real number, then the value obtained by replacing $x$ by a in $p (x)$ is called value of $p (x)$ at $x = a$ and is denoted by $p (a)$ . For example : Find the value of $p (x) = x^{3} - 6 x^{2} + 11 x - 6$ at $x = - 2$ $\Rightarrow p (- 2) = (- 2)^{3} - 6 (- 2)^{2} + 11 (- 2) - 6 = - 8 - 24 - 22 - 6$ $\Rightarrow p (- 2) = - 60$ Zero of a polynomial : A real number $a$ is a zero of the polynomial $p (x)$ if $p (a) = 0$ . For example: Consider $p (x) = x^{3} - 6 x^{2} + 11 x - 6$ $p (1) = (1)^{3} - 6 (1)^{2} + 11 (1) - 6 = 1 - 6 + 11 - 6 = 0$ $p (2) = (2)^{3} - 6 (2)^{2} + 11 (2) - 6 = 8 - 24 + 22 - 6 = 0$ $p (3) = (3)^{3} - 6 (3)^{2} + 11 (3) - 6 = 27 - 54 + 33 - 6 = 0$ Thus, 1,2 and 3 are called the zeros of polynomial $p (x)$ .

3.0Geometrical meaning of the zeros of a polynomial

Geometrically the zeros of a polynomials $f (x)$ are the x -co-ordinates of the points where the graph $y = f (x)$ intersects $x$ -axis. To understand it, we will see the geometrical representations of linear and quadratic polynomials. Geometrical representation of the zero of a linear polynomial $y = a x + b$

The graph of $y = a x + b$ intersects the $x$ -axis at only one point $A$ . Therefore, we conclude that the linear polynomial $a x + b$ has one and only one zero. Geometrical representation of the zero of a quadratic polynomial $y = a x^{2} + b x + c$

Consider the following cases:

Case-I : Here, the graph cuts x -axis at two distinct points A and A'. (see figure)

The $x$ -coordinates of $A$ and $A^{'}$ are two zeros of the quadratic polynomial $a x^{2} + b x + c$

Case-II : Here, the graph touches the x -axis at exactly one point (see fig). So, the two points A and A' of Case-I coincide here to become one point A. The x -coordinate of A is the only zero for the quadratic polynomial $ax^{2} + bx + c$ in this case.

Case-III : Here, the graph is either completely above the x -axis or completely below the x -axis (see fig.). So, it does not cut the x -axis at any point. So, the quadratic polynomial $a x^{2} + b x + c$ has no zero in this case.

So, you can see geometrically that a quadratic polynomial can have either two distinct zeros or one zero, or no zero. This also means that a polynomial of degree 2 has at most two zeros.

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 zeros.

Building Concepts 1 The graphical representation of $y = a x^{2} + b x + c$ is shown in the given figure. Does it have (i) any solution (ii) zero(s)

Explanation: (i) It has infinite solutions because there are infinite points which satisfy the equation

y = a x^{2} + b x + c

. (ii) It has no zero because there is no real x for which y gets zero or graphically there is no point of intersection of

x

-axis and the curve.

Building Concepts 2 Look at the graphs given below. Each is the graph of $y = p (x)$ , where $p (x)$ is a polynomial. For each of the graphs, find the number of zeros of $p (x)$ .

Explanation: (i) The number of zeros is 1 as the graph intersects the x -axis at one point only. (ii) The number of zeros is 2 as the graph intersects the x -axis at two points. (iii) The number of zeros is 3 as the graph intersects the

x

-axis at three point only. (iv) The number of zeros is 1 as the graph intersects the x -axis at one point only. (v) The number of zeros is 1 as the graph intersects the

x

-axis at one point only. (vi) The number of zeros is 4 as the graph intersects the

x

-axis at four point only.

4.0Relationship between the zeros and coefficients of a polynomial

For a linear polynomial $a x + b, (a \neq = 0)$ , we have, zero of a linear polynomial $= - \frac{b}{a} = - \frac{( constant term )}{( coefficient of x )}$ For a quadratic polynomial $a x^{2} + b x + c (a \neq = 0)$ , with $α$ and $β$ as it's zeros, $(x - α)$ and $(x - β)$ are the factors of $a x^{2} + b x + c$ .

Therefore, $ax^{2} + bx + c = K (x - α) (x - β)$ , (where $K$ is a constant to balance the equation of the coefficient of $x^{2}$ i.e. $a \neq = 1$ .) $= K x^{2} - K (α + β) x + K α β$ comparing the coefficients of $x^{2}, x$ and constant terms on both the sides, we get $a = K, b = - K (α + β)$ and $c = K α β$

This gives Sum of zeros $x$ Product of zeros $= α β = \frac{c}{a} = \frac{(constant term)}{(coefficient of x )}$ If $α$ and $β$ are the zeros of a quadratic polynomial $f (x)$ . Then polynomial $f (x)$ is given by $f (x) = K {x^{2} - (a + b) x + ab}$ or $β$ sum of the zeros $) x +$ product of the zeros $}$

The cubic polynomial whose zeros are $α, β$ and $γ$ is given by $ax^{2} + bx + c = K (x - α) (x - β)$ $x^{2} + (α β + β γ + γ α) x -$ $α β γ}$

If a polynomial has no real zero, it does not mean that it has no solution.

Numerical Ability 1 Find the zeros of the quadratic polynomial $x^{2} + 7 x + 12$ and verify the relation between the zeros and its coefficients. Solution: We have, $f (x) = x^{2} + 7 x + 12 = x^{2} + 4 x + 3 x + 12 \Rightarrow f (x) = x (x + 4) + 3 (x + 4) \Rightarrow f (x) = (x + 4) (x + 3) The zeros of f (x) are given by f (x) = 0 \Rightarrow x^{2} + 7 x + 12 = 0 \Rightarrow (x + 4) (x + 3) = 0 \Rightarrow x + 4 = 0 or x + 3 = 0 \Rightarrow x = - 4 or x = - 3$ For a cubic polynomial $a x^{3} + b x^{2} + c x + d (a \neq = 0)$ with $α, β$ and $γ$ as its zeros, we have $α + β + γ = \frac{- b}{a}$ $α β + β γ + γ α = c / a$ $α β γ = - d / a$ Thus, the zeros of $f (x) = x^{2} + 7 x + 12$ are $f (x) = x^{2} + 7 x + 12 = x^{2} + 4 x + 3 x + 12 \Rightarrow f (x) = x (x + 4) + 3 (x + 4) \Rightarrow f (x) = (x + 4) (x + 3) The zeros of f (x) are given by f (x) = 0 \Rightarrow x^{2} + 7 x + 12 = 0 \Rightarrow (x + 4) (x + 3) = 0 \Rightarrow x + 4 = 0 or x + 3 = 0 \Rightarrow x = - 4 or x = - 3$ and $β = - 3$ Now,sum of the zeros $= α + β = (- 4) + (- 3) = - 7$ and $- \frac{Coefficient of x}{Coefficient of x ^{2}} = - \frac{7}{1} = - 7$ $∴$ Sum of the zeros $= - \frac{Coefficient of x}{Coefficient of x ^{2}}$ Product of the zeros $= α β = (- 4) \times (- 3) = 12$ and, $\frac{Constant term}{Coefficient of x ^{2}} = \frac{12}{1} = 12$ $∴$ Product of the zeros $= \frac{Constant term}{Coefficient of x ^{2}}$

Numerical Ability 2 Find the zeros of the quadratic polynomial $f (x) = ab x^{2} + (b^{2} + a c) x + b c$ and verify the relationship between the zeros and its coefficients. Solution: $f (x) = ab x^{2} + (b^{2} + a c) x + b c = ab x^{2} + b^{2} x + a c x + b c$ $= b x (a x + b) + c (a x + b) = (a x + b) (b x + c)$ So, the value of $f (x)$ is zero when $a x + b = 0$ or $b x + c = 0$ , i.e. $x = \frac{- b}{a}$ or $x = \frac{- c}{b}$ Therefore, $\frac{- b}{a}$ and $\frac{- c}{b}$ are the zeros of $f (x)$ . Now, sum of zeros $= (\frac{- b}{a}) + (\frac{- c}{b}) = \frac{- b ^{2} - ac}{ab}$ $= \frac{- ( b ^{2} + a c )}{ab} = \frac{Coefficient of x}{Coefficient of x ^{2}}$ Product of zeros $= (\frac{- b}{a}) (\frac{- c}{b}) = \frac{b c}{ab} = \frac{Constant term}{Coefficientof x ^{2}}$

Numerical Ability 3 Find a quadratic polynomial whose zeros are 5 and -2 respectively. Solution: Let the quadratic polynomial be $ax^{2} + bx + c$ , and it's zeros be $α$ and $β$ we have, $α + β = 5 + (- 2) = 3$ $α β = 5 (- 2) = - 10$ We know that a quadratic polynomial when the sum and product of its zeros is given by $f (x) = K [x^{2} - ($ sum of zeros $) x +$ product of zeros] where $K$ is a constant so, $f (x) = K {x^{2} - 3 x - 10}$

Numerical Ability 4 Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively: (i) $\frac{1}{4}, - 1$ (ii) $2, \frac{1}{3}$ We know that a quadratic polynomial when the sum and product of its zeros are given by $f (x) = k {x^{2} -$ (Sum of the zeros) $x +$ Product of the zeros $}$ , where $k$ is a constant. (i) Required quadratic polynomial $f (x)$ is given by $f (x) = k {x^{2} - \frac{1}{4} x - 1}$ (ii) Required quadratic polynomial $f (x)$ is given by $f (x) = k (x^{2} - 2 x + \frac{1}{3})$

5.0Division Algorithm for polynomials

If $f (x)$ is a polynomial and $g (x)$ is a non-zero polynomial, then there exist two polynomials $q (x)$ and $r (x)$ such that $f (x) = g (x) \times q (x) + r (x)$ , where $r (x) = 0$ or degree $r (x) <$ degree $g (x)$ . In other words,

Dividend = Divisor \times Quotient + Remainder

Remark : If $r (x) = 0$ , then polynomial $g (x)$ is a factor of polynomial $f (x)$ .

Building Concepts 3 Check whether the polynomial $m^{2} - 4$ is a factor of the polynomial $m^{4} + 9 m^{3} - 36 m - 16$ , by dividing the second polynomial by the first polynomial. Explanation: We have

Since the remainder is zero, therefore, the polynomial

m^{2} - 4

is a factor of the polynomial

m^{4} + 9 m^{3} - 36 m - 16

.

Building Concepts 4 Find all the zeros of $2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2$ , if you know that two of its zeros are $2$ and $- 2$ . Explanation: Let $p (x) = 2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2$ be the given polynomial. Since two zeros are $2$ and $- 2$ so, $(x - 2)$ and $(x + 2)$ are both factors of the given polynomial $p (x)$ . Also, $(x - 2) (x + 2) = (x^{2} - 2)$ is a factor of the polynomial $p (x)$ . Now, we divide the given polynomial by $x^{2} - 2$ .

By division algorithm, we have

2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2 = (x^{2} - 2) (2 x^{2} - 3 x + 1) \Rightarrow 2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2 = (x - 2) (x + 2) (2 x^{2} - 2 x - x + 1) \Rightarrow 2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2 = (x - 2) (x + 2) {2 x (x - 1) - (x - 1)} \Rightarrow 2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2 = (x - 2) (x + 2) (x - 1) (2 x - 1)

When

p (x) = 0, x = 2, - 2, 1, \frac{1}{2}

Hence, all the zeros of the polynomial

2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2

are

2, - 2, 1

and

\frac{1}{2}

Numerical Ability 5 Divide the polynomial $2 x^{2} + 3 x + 1$ by the polynomial $x + 2$ and verify the division algorithm. Solution: We have

Clearly, quotient

= 2 x - 1

and remainder

= 3

Also,

(x + 2) (2 x - 1) + 3

= 2 x^{2} + 4 x - x - 2 + 3

= 2 x^{2} + 3 x + 1

i.e.,

2 x^{2} + 3 x + 1 = (x + 2) (2 x - 1) + 3

. Thus, Dividend

=

Divisor

\times

Quotient + Remainder .

6.0Symmetric functions of the zeros

Let $α, β$ be the zeros of a quadratic polynomial, then the expression of the form $α + β$ ; $(α^{2} + β^{2}); α β$ are called the symmetric functions of the zeros. By symmetric function we mean that the function remains invariant (unaltered) in values when the roots are changed cyclically. In other words, an expression involving $α$ and $β$ which remains unchanged by interchanging $α$ and $β$ is called a symmetric function of $α$ and $β$ .

Some useful relations involving $α$ and $β$ are :- (i) $α^{2} + β^{2} = (α + β)^{2} - 2 α β$ (ii) $(α - β)^{2} = (α + β)^{2} - 4 α β$ (iii) $α^{2} - β^{2} = (α + β) (α - β) = (α + β) (α + β)^{2} - 4 α β$ (iv) $α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β)$ (v) $α^{3} - β^{3} = (α - β)^{3} + 3 α β (α - β)$

$(a + b)^{2} = a^{2} + b^{2} + 2 ab$
$(a - b)^{2} = a^{2} + b^{2} - 2 ab$
$(a + b)^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}$
$(a - b)^{3} = a^{3} - b^{3} - 3 a^{2} b + 3 a b^{2}$

Numerical Ability 6 If $α$ and $β$ are the zeros of the quadratic polynomial $f (x) = a x^{2} + b x + c$ then calculate (i) $α^{2} + β^{2}$ (ii) $\frac{α ^{2}}{β} + \frac{β ^{2}}{α}$ Solution Since $α$ and $β$ are the zeros of the quadratic polynomial $f (x) = a^{2} + b x + c ∴ α + β = - \frac{b}{a} and α β = \frac{c}{a}$ (i) We have, $α^{2} + β^{2} = (α + β)^{2} - 2 α β$ $\Rightarrow α^{2} + β^{2} = (\frac{- b}{a})^{2} - \frac{2 c}{a} = \frac{b ^{2} - 2 a c}{a ^{2}}$ (ii) We have, $\frac{α ^{2}}{β} + \frac{β ^{2}}{α} = \frac{α ^{3} + β ^{3}}{α β} = \frac{( α + β ) ^{3} - 3 α β ( α + β )}{α β} = \frac{( - \frac{b}{a} ) ^{3} - 3 ( \frac{c}{a} ) ( - \frac{b}{a} )}{\frac{c}{a}} \Rightarrow \frac{α ^{2}}{β} + \frac{β ^{2}}{α} = \frac{3 ab c - b ^{3}}{a ^{2} c}$

Numerical Ability 7 If $α$ and $β$ are the zeros of the quadratic polynomial $p (s) = 3 s^{2} - 6 s + 4$ , find the value of $\frac{α}{β} + \frac{β}{α} + 2 (\frac{1}{α} + \frac{1}{β}) + 3 α β$ Solution: Since $α$ and $β$ are the zeros of the polynomial $p (s) = 3 s^{2} - 6 s + 4$ . $∴ α + β = \frac{- ( - 6 )}{3} = 2$ and $α β = \frac{4}{3}$ We have $\frac{α}{β} + \frac{β}{α} + 2 (\frac{1}{α} + \frac{1}{β}) + 3 α β = \frac{α ^{2} + β ^{2}}{α β} + 2 (\frac{β + α}{α β}) + 3 α β$ $= \frac{( α + β ) ^{2} - 2 α β}{α β} + 2 \frac{( α + β )}{α β} + 3 α β = \frac{( 2 ) ^{2} - 2 \times \frac{4}{3}}{\frac{4}{3}} + \frac{2 \times 2}{\frac{4}{3}} + 3 \times \frac{4}{3} = 8$

Numerical Ability 8 If $α$ and $β$ are the roots (zeros) of the polynomial $f (x) = x^{2} - 3 x + k$ such that $α - β = 1$ , find the value of $k$ . Solution: Since $α$ and $β$ are the roots (zeros) of the polynomial $f (x) = x^{2} - 3 x + k$ . $∴ α + β = \frac{- ( - 3 )}{1} = 3 and α β = k . We have α - β = 1 \Rightarrow (α - β)^{2} = (1)^{2} \Rightarrow α^{2} - 2 α β + β^{2} = 1 \Rightarrow (α^{2} + β^{2}) - 2 α β = 1 \Rightarrow {(α + β)^{2} - 2 α β} - 2 α β = 1 \Rightarrow (α + β)^{2} - 4 α β = 1 \Rightarrow (3)^{2} - 4 \times k = 1 \Rightarrow 9 - 4 k = 1 \Rightarrow 4 k = 8 \Rightarrow k = 2$ Hence, the value of k is 2 .

Numerical Ability 9 If $α, β$ are the zeros of the polynomial $f (x) = 2 x^{2} + 5 x + k$ satisfying the relation $α^{2} + β^{2} + α β = \frac{21}{4}$ , then find the value of $k$ for this to be possible. Solution Since $α$ and $β$ are the zeros of the polynomial $f (x) = 2 x^{2} + 5 x + k$ . $∴ α + β = \frac{- 5}{2}$ and $α β = \frac{k}{2}$ Now, $α^{2} + β^{2} + α β = \frac{21}{4}$ $\Rightarrow (α^{2} + β^{2} + 2 α β) - α β = \frac{21}{4}$ $\Rightarrow (α + β)^{2} - α β = \frac{21}{4}$ $\Rightarrow \frac{25}{4} - \frac{k}{2} = \frac{21}{4}$ $[∴ α + β = - \frac{5}{2}$ and $α β = \frac{k}{2}]$ $\Rightarrow - \frac{k}{2} = - 1$ $\Rightarrow k = 2$

7.0Memory Map