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Polynomials

An algebraic expression of the form p(x)=an​xn+an−1​xn−1+an−2​xn−2+…+a1​x1+a0​x0, where (i) an​=0 (ii) a0​,a1​,a2​,…an​ are real numbers (iii) power of x is a non negative integer, is called a polynomial.

Where an​,an−1​,an−2​, are coefficients of xn,xn−1 x0 respectively and an​xn,an−1​xn−1,an−2​xn−2,…..a0​x0 are terms of the polynomial. Here the term an​xn is called the leading term and its coefficient an​ is the leading coefficient. For e.g.: p(u)=21​u3−3u2+2u−4 is a polynomial in variable u. 21​u3,−3u2,2u and -4 are known as terms of polynomial and 21​,−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 variablesNameExample
1Univariatex+9
2Bivariatex+y+9
3Trivariatex+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 =x2y3+x7y3+x2y2 here, degrees of term x2y3,x7y3,x2y2 are 5,10,4 respectively. Degree of given polynomial =10 Polynomials classified by degree
DegreeNameExample
−∞Zero0
0(non-zero) Constant1
1Linearx+1
2Quadraticx2+1
3Cubicx3+2

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 termsNameExample
0Zero polynomial0
1Monomialx2
2Binomialx2+1
3Trinomialx2+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)=x3−6x2+11x−6 at x=−2 ⇒p(−2)=(−2)3−6(−2)2+11(−2)−6=−8−24−22−6 ⇒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)=x3−6x2+11x−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=ax+b

The graph of y=ax+b intersects the x-axis at only one point A. Therefore, we conclude that the linear polynomial ax+b has one and only one zero. Geometrical representation of the zero of a quadratic polynomial y=ax2+bx+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 ax2+bx+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 ax2+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 ax2+bx+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=ax2+bx+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=ax2+bx+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 ax+b,(a=0), we have, zero of a linear polynomial =−ab​=−(coefficient of x)( constant term )​ For a quadratic polynomial ax2+bx+c(a=0), with α and β as it's zeros, (x−α) and (x−β) are the factors of ax2+bx+c.

Therefore, ax2+bx+c=K(x−α)(x−β), (where K is a constant to balance the equation of the coefficient of x2 i.e. a=1.) =Kx2−K(α+β)x+Kαβ comparing the coefficients of x2,x and constant terms on both the sides, we get a=K,b=−K(α+β) and c=Kαβ

This gives Sum of zeros =α+β=−ab​=−( coefficient of x2)( coefficient of x)​ Product of zeros =αβ=ac​= (coefficient of x )  (constant term) ​ If α and β are the zeros of a quadratic polynomial f(x). Then polynomial f(x) is given by f(x)=K{x2−(a+b)x+ab} or f(x)=K{x2−( sum of the zeros )x+ product of the zeros }

The cubic polynomial whose zeros are α,β and γ is given by f(x)=k{x3−(α+β+γ) x2+(αβ+βγ+γα)x− αβγ}

  • 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 x2+7x+12 and verify the relation between the zeros and its coefficients. Solution: We have, f(x)=x2+7x+12=x2+4x+3x+12⇒f(x)=x(x+4)+3(x+4)⇒f(x)=(x+4)(x+3) The zeros of f(x) are given by f(x)=0⇒x2+7x+12=0⇒(x+4)(x+3)=0⇒x+4=0 or x+3=0⇒x=−4 or x=−3 For a cubic polynomial ax3+bx2+cx+d(a=0) with α,β and γ as its zeros, we have α+β+γ=a−b​ αβ+βγ+γα=c/a αβγ=−d/a Thus, the zeros of f(x)=x2+7x+12 are α=−4 and β=−3 Now,sum of the zeros =α+β=(−4)+(−3)=−7 and − Coefficient of x2 Coefficient of x​=−17​=−7 ∴ Sum of the zeros =− Coefficient of x2 Coefficient of x​ Product of the zeros =αβ=(−4)×(−3)=12 and,  Coefficient of x2 Constant term ​=112​=12 ∴ Product of the zeros = Coefficient of x2 Constant term ​

Numerical Ability 2 Find the zeros of the quadratic polynomial f(x)=abx2+(b2+ac)x+bc and verify the relationship between the zeros and its coefficients. Solution: f(x)=abx2+(b2+ac)x+bc=abx2+b2x+acx+bc =bx(ax+b)+c(ax+b)=(ax+b)(bx+c) So, the value of f(x) is zero when ax+b=0 or bx+c=0, i.e. x=a−b​ or x=b−c​ Therefore, a−b​ and b−c​ are the zeros of f(x). Now, sum of zeros =(a−b​)+(b−c​)=ab−b2−ac​ =ab−(b2+ac)​= Coefficient of x2 Coefficient of x​ Product of zeros =(a−b​)(b−c​)=abbc​= Coefficientof x2 Constant term ​

Numerical Ability 3 Find a quadratic polynomial whose zeros are 5 and -2 respectively. Solution: Let the quadratic polynomial be ax2+bx+c, and it's zeros be α and β we have, α+β=5+(−2)=3 αβ=5(−2)=−10 We know that a quadratic polynomial when the sum and product of its zeros is given by f(x)=K[x2−( sum of zeros )x+ product of zeros] where K is a constant so, f(x)=K{x2−3x−10}

Numerical Ability 4 Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively: (i) 41​,−1 (ii) 2​,31​ We know that a quadratic polynomial when the sum and product of its zeros are given by f(x)=k{x2− (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{x2−41​x−1} (ii) Required quadratic polynomial f(x) is given by f(x)=k(x2−2​x+31​)

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)×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 m2−4 is a factor of the polynomial m4+9m3−36m−16, by dividing the second polynomial by the first polynomial. Explanation: We have

Since the remainder is zero, therefore, the polynomial m2−4 is a factor of the polynomial m4+9m3−36m−16.

Building Concepts 4 Find all the zeros of 2x4−3x3−3x2+6x−2, if you know that two of its zeros are 2​ and −2​. Explanation: Let p(x)=2x4−3x3−3x2+6x−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​)=(x2−2) is a factor of the polynomial p(x). Now, we divide the given polynomial by x2−2.

By division algorithm, we have 2x4−3x3−3x2+6x−2=(x2−2)(2x2−3x+1)⇒2x4−3x3−3x2+6x−2=(x−2​)(x+2​)(2x2−2x−x+1)⇒2x4−3x3−3x2+6x−2=(x−2​)(x+2​){2x(x−1)−(x−1)}⇒2x4−3x3−3x2+6x−2=(x−2​)(x+2​)(x−1)(2x−1) When p(x)=0,x=2​,−2​,1,21​ Hence, all the zeros of the polynomial 2x4−3x3−3x2+6x−2 are 2​,−2​,1 and 21​

Numerical Ability 5 Divide the polynomial 2x2+3x+1 by the polynomial x+2 and verify the division algorithm. Solution: We have

Clearly, quotient =2x−1 and remainder =3 Also, (x+2)(2x−1)+3 =2x2+4x−x−2+3 =2x2+3x+1 i.e., 2x2+3x+1=(x+2)(2x−1)+3. Thus, Dividend = Divisor × 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; (α2+β2);αβ 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) α2+β2=(α+β)2−2αβ (ii) (α−β)2=(α+β)2−4αβ (iii) α2−β2=(α+β)(α−β)=(α+β)(α+β)2−4αβ​ (iv) α3+β3=(α+β)3−3αβ(α+β) (v) α3−β3=(α−β)3+3αβ(α−β)

  • (a+b)2=a2+b2+2ab
  • (a−b)2=a2+b2−2ab
  • (a+b)3=a3+b3+3a2b+3ab2
  • (a−b)3=a3−b3−3a2b+3ab2

Numerical Ability 6 If α and β are the zeros of the quadratic polynomial f(x)=ax2+bx+c then calculate (i) α2+β2 (ii) βα2​+αβ2​ Solution Since α and β are the zeros of the quadratic polynomial f(x)=a2+bx+c∴α+β=−ab​ and αβ=ac​ (i) We have, α2+β2=(α+β)2−2αβ ⇒α2+β2=(a−b​)2−a2c​=a2b2−2ac​ (ii) We have, βα2​+αβ2​=αβα3+β3​=αβ(α+β)3−3αβ(α+β)​=ac​(−ab​)3−3(ac​)(−ab​)​⇒βα2​+αβ2​=a2c3abc−b3​

Numerical Ability 7 If α and β are the zeros of the quadratic polynomial p(s)=3s2−6s+4, find the value of βα​+αβ​+2(α1​+β1​)+3αβ Solution: Since α and β are the zeros of the polynomial p(s)=3s2−6s+4. ∴α+β=3−(−6)​=2 and αβ=34​ We have βα​+αβ​+2(α1​+β1​)+3αβ=αβα2+β2​+2(αββ+α​)+3αβ =αβ(α+β)2−2αβ​+2αβ(α+β)​+3αβ=34​(2)2−2×34​​+34​2×2​+3×34​=8

Numerical Ability 8 If α and β are the roots (zeros) of the polynomial f(x)=x2−3x+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)=x2−3x+k. ∴α+β=1−(−3)​=3 and αβ=k. We have α−β=1⇒(α−β)2=(1)2⇒α2−2αβ+β2=1⇒(α2+β2)−2αβ=1⇒{(α+β)2−2αβ}−2αβ=1⇒(α+β)2−4αβ=1⇒(3)2−4×k=1⇒9−4k=1⇒4k=8⇒k=2 Hence, the value of k is 2 .

Numerical Ability 9 If α,β are the zeros of the polynomial f(x)=2x2+5x+k satisfying the relation α2+β2+αβ=421​, then find the value of k for this to be possible. Solution Since α and β are the zeros of the polynomial f(x)=2x2+5x+k. ∴α+β=2−5​ and αβ=2k​ Now, α2+β2+αβ=421​ ⇒(α2+β2+2αβ)−αβ=421​ ⇒(α+β)2−αβ=421​ ⇒425​−2k​=421​ [∴α+β=−25​ and αβ=2k​] ⇒−2k​=−1 ⇒k=2

7.0Memory Map

On this page


  • 1.0Types of polynomials:
  • 1.1Polynomials classified by number of distinct variables
  • 2.0Value of a polynomial
  • 3.0Geometrical meaning of the zeros of a polynomial
  • 4.0Relationship between the zeros and coefficients of a polynomial
  • 5.0Division Algorithm for polynomials
  • 6.0Symmetric functions of the zeros
  • 7.0Memory Map

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