Class 10th Chapter-2 Polynomial Revision Notes & Solution

Class 10th Chapter-2 Polynomial Revision Notes & Solution

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Algebraic Expressions An algebraic expression is an expression made up of variables and constants along with mathematical operators. An algebraic expression is a sum of terms, which are considered to be building blocks for expressions. A term is a product of variables and constants. A term can be an algebraic expression in itself.

Examples of a term – 3 which is just a constant. – 2x, which is the product of constant ‘2’ and the variable ‘x’ – 4xy, which is the product of the constant ‘4’ and the variables ‘x’ and ‘y’. – 5x2y, which is the product of 5, x, x and y.

The constant in each term is referred to as the coefficient. Example of an algebraic expression: 3x2y+4xy+5x+6 which is the sum of four terms: 3x2y, 4xy, 5x and 6. An algebraic expression can have any number of terms. The coefficient in each term can be any real number.

There can be any number of variables in an algebraic expression. The exponent on the variables, however, must be rational numbers.

Polynomial An algebraic expression can have exponents that are rational numbers. However, a polynomial is an algebraic

expression in which the exponent on any variable is a whole number. 5x3+3x+1 is an example of a polynomial. It is an algebraic expression as well. 2x+3√x is an algebraic expression, but not a polynomial. – since the exponent on x is 1/2 which is not a whole

number.

Degree of a Polynomial For a polynomial in one variable – the highest exponent on the variable in a polynomial is the degree of the

polynomial. Example: The degree of the polynomial x2+2x+3 is 2, as the highest power of x in the given expression is x2.

Types Of Polynomials Polynomials can be classified based on:

a) Number of terms b) Degree of the polynomial.

Types of polynomials based on the number of terms a) Monomial – A polynomial with just one term. Example: 2x, 6x2, 9xy b) Binomial – A polynomial with two terms. Example: 4x2+x, 5x+4 a) Trinomial – A polynomial with three terms. Example: x2+3x+4 Types of Polynomials based on Degree Linear Polynomial A polynomial whose degree is one is called a linear polynomial.

For example, 2x+1 is a linear polynomial. Quadratic Polynomial A polynomial of degree two is called a quadratic polynomial.

For example, 3x2+8x+5 is a quadratic polynomial. Cubic Polynomial A polynomial of degree three is called a cubic polynomial.

For example, 2x3+5x2+9x+15 is a cubic polynomial.



Graphical Representations Let us learn here how to represent polynomial equation on the graph. Representing Equations on a Graph Any equation can be represented as a graph on the Cartesian plane, where each point on the graph represents the

x and y coordinates of the point that satisfies the equation. An equation can be seen as a constraint placed on the x and y coordinates of a point, and any point that satisfies that constraint will lie on the curve

For example, the equation y = x, on a graph, will be a straight line that joins all the points which have their x coordinate equal to their y coordinate. Example – (1,1), (2,2) and so on.

Geometrical Representation of a Linear Polynomial The graph of a linear polynomial is a straight line. It cuts the X-axis at exactly one point.



Linear graph Geometrical Representation of a Quadratic Polynomial

The graph of a quadratic polynomial is a parabola It looks like a U which either opens upwards or opens downwards depending on the value of ‘a’ in ax2+bx+c If ‘a’ is positive, then parabola opens upwards and if ‘a’ is negative then it opens downwards It can cut the x-axis at 0, 1 or two points

Graph of a polynomial which cuts the x-axis in two distinct points (a>0)

Graph of a Quadratic polynomial which touches the x-axis at one point (a>0)



Graph of a Quadratic polynomial that doesn’t touch the x-axis (a<0) Graph of the polynomial x^n For a polynomial of the form y=xn where n is a whole number:

as n increases, the graph becomes steeper or draws closer to the Y-axis If n is odd, the graph lies in the first and third quadrants If n is even, the graph lies in the first and second quadrants The graph of y=−xn is the reflection of the graph of y=xn on the x-axis

Graph of polynomials with different degrees.



Zeroes of a Polynomial A zero of a polynomial p(x) is the value of x for which the value of p(x) is 0. If k is a zero of p(x), then p(k)=0.

For example, consider a polynomial p(x)=x2−3x+2. When x=1, the value of p(x) will be equal to p(1)=12−3×1+2 =1−3+2 =0

Since p(x)=0 at x=1, we say that 1 is a zero of the polynomial x2−3x+2

Geometrical Meaning of Zeros of a Polynomial Geometrically, zeros of a polynomial are the points where its graph cuts the x-axis.

(i) One zero (ii) Two zeros (iii) Three zeros Here A, B and C correspond to the zeros of the polynomial represented by the graphs.

Number of Zeros In general, a polynomial of degree n has at most n zeros. 1. A linear polynomial has one zero, 2. A quadratic polynomial has at most two zeros. 3. A cubic polynomial has at most 3 zeros.

Factorisation of Polynomials Quadratic polynomials can be factorized by splitting the middle term. For example, consider the polynomial 2x2−5x+3 Splitting the middle term: The middle term in the polynomial 2x2−5x+3 is -5x. This must be expressed as a sum of two terms such that the

product of their coefficients is equal to the product of 2 and 3 (coefficient of x2 and the constant term) −5 can be expressed as (−2)+(−3), as −2×−3=6=2×3 Thus, 2x2−5x+3=2x2−2x−3x+3 Now, identify the common factors in individual groups 2x2−2x−3x+3=2x(x−1)−3(x−1) Taking (x−1) as the common factor, this can be expressed as: 2x(x−1)−3(x−1)=(x−1)(2x−3)



Relationship between Zeroes and Coefficients of a Polynomial For Quadratic Polynomial:

If α and β are the roots of a quadratic polynomial ax2+bx+c, then, α + β = -b/a Sum of zeroes = -coefficient of x /coefficient of x2 αβ = c/a Product of zeroes = constant term / coefficient of x2 For Cubic Polynomial If α,β and γ are the roots of a cubic polynomial ax3+bx2+cx+d, then α+β+γ = -b/a αβ +βγ +γα = c/a αβγ = -d/a

Division Algorithm To divide one polynomial by another, follow the steps given below. Step 1: arrange the terms of the dividend and the divisor in the decreasing order of their degrees. Step 2: To obtain the first term of the quotient, divide the highest degree term of the dividend by the highest

degree term of the divisor Then carry out the division process. Step 3: The remainder from the previous division becomes the dividend for the next step. Repeat this process until

the degree of the remainder is less than the degree of the divisor.

Algebraic Identities 1. (a+b)2=a2+2ab+b2

2. (a−b)2=a2−2ab+b2 3. (x+a)(x+b)=x2+(a+b)x+ab 4. a2−b2=(a+b)(a−b) 5. a3−b3=(a−b)(a2+ab+b2) 6. a3+b3=(a+b)(a2−ab+b2) 7. (a+b)3=a3+3a2b+3ab2+b3 8. (a−b)3=a3−3a2b+3ab2−b3



Access Answers to NCERT Class 10 Maths Chapter 2 – Polynomials Exercise 2.1 Page: 28

1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Solutions: Graphical method to find zeroes:- Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis. (i) In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does not cut it at any

point. (ii) In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at only one point. (iii) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at any three points. (iv) In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at two points. (v) In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at four points. (vi) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at three points.



Exercise 2.2 Page: 33 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the

coefficients. Solutions: (i) x2–2x –8 ⇒x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2) Therefore, zeroes of polynomial equation x2–2x–8 are (4, -2) Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x2) Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x2) (ii) 4s2–4s+1 ⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1) Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2) Sum of zeroes = (½)+(1/2) = 1 = -4/4 = -(Coefficient of s)/(Coefficient of s2) Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 ) (iii) 6x2–3–7x ⇒6x2–7x–3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3) Therefore, zeroes of polynomial equation 6x2–3–7x are (-1/3, 3/2) Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x2) Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x2 ) (iv) 4u2+8u ⇒ 4u(u+2) Therefore, zeroes of polynomial equation 4u2 + 8u are (0, -2). Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u2) Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u2 ) (v) t2–15 ⇒ t2 = 15 or t = ±√15 Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15) Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t2) Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 ) (vi) 3x2–x–4 ⇒ 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1) Therefore, zeroes of polynomial equation3x2 – x – 4 are (4/3, -1) Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x2) Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x2 ) 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (i) 1/4 , -1 Solution: From the formulas of sum and product of zeroes, we know, Sum of zeroes = α+β Product of zeroes = α β Sum of zeroes = α+β = 1/4 Product of zeroes = α β = -1 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly

as:- x2–(α+β)x +αβ = 0 x2–(1/4)x +(-1) = 0



4x2–x-4 = 0 Thus,4x2–x–4 is the quadratic polynomial. (ii)√2, 1/3 Solution: Sum of zeroes = α + β =√2 Product of zeroes = α β = 1/3 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly

as:- x2–(α+β)x +αβ = 0 x2 –(√2)x + (1/3) = 0 3x2-3√2x+1 = 0 Thus, 3x2-3√2x+1 is the quadratic polynomial. (iii) 0, √5 Solution: Given, Sum of zeroes = α+β = 0 Product of zeroes = α β = √5 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:- x2–(α+β)x +αβ = 0 x2–(0)x +√5= 0 Thus, x2+√5 is the quadratic polynomial. (iv) 1, 1 Solution: Given, Sum of zeroes = α+β = 1 Product of zeroes = α β = 1 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly

as:- x2–(α+β)x +αβ = 0 x2–x+1 = 0 Thus , x2–x+1is the quadratic polynomial. (v) -1/4, 1/4 Solution: Given, Sum of zeroes = α+β = -1/4 Product of zeroes = α β = 1/4 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly

as:- x2–(α+β)x +αβ = 0 x2–(-1/4)x +(1/4) = 0 4x2+x+1 = 0 Thus,4x2+x+1 is the quadratic polynomial.



(vi) 4, 1 Solution: Given, Sum of zeroes = α+β = Product of zeroes = αβ = 1 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly

as:- x2–(α+β)x+αβ = 0 x2–4x+1 = 0 Thus, x2–4x+1 is the quadratic polynomial.

Exercise 2.3 Page: 36

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x3-3x2+5x–3 , g(x) = x2–2 Solution: Given, Dividend = p(x) = x3-3x2+5x–3 Divisor = g(x) = x2– 2

Therefore, upon division we get, Quotient = x–3 Remainder = 7x–9 (ii) p(x) = x4-3x2+4x+5 , g(x) = x2+1-x Solution: Given, Dividend = p(x) = x4 – 3x2 + 4x +5 Divisor = g(x) = x2 +1-x



Therefore, upon division we get, Quotient = x2 + x–3 Remainder = 8 (iii) p(x) =x4–5x+6, g(x) = 2–x2 Solution: Given, Dividend = p(x) =x4 – 5x + 6 = x4 +0x2–5x+6 Divisor = g(x) = 2–x2 = –x2+2

Therefore, upon division we get, Quotient = -x2-2 Remainder = -5x + 10 2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by

the first polynomial: (i) t2-3, 2t4 +3t3-2t2-9t-12 Solutions: Given, First polynomial = t2-3 Second polynomial = 2t4 +3t3-2t2 -9t-12



As we can see, the remainder is left as 0. Therefore, we say that, t2-3 is a factor of 2t2+3t+4. (ii)x2+3x+1 , 3x4+5x3-7x2+2x+2 Solutions: Given, First polynomial = x2+3x+1 Second polynomial = 3x4+5x3-7x2+2x+2

As we can see, the remainder is left as 0. Therefore, we say that, x2 + 3x + 1 is a factor of 3x4+5x3-7x2+2x+2. (iii) x3-3x+1, x5-4x3+x2+3x+1 Solutions: Given, First polynomial = x3-3x+1 Second polynomial = x5-4x3+x2+3x+1

As we can see, the remainder is not equal to 0. Therefore, we say that, x3-3x+1 is not a factor of x5-4x3+x2+3x+1 .



3. Obtain all other zeroes of 3x4+6x3-2x2-10x-5, if two of its zeroes are √(5/3) and – √(5/3). Solutions: Since this is a polynomial equation of degree 4, hence there will be total 4 roots. √(5/3) and – √(5/3) are zeroes of polynomial f(x). ∴ (x –√(5/3)) (x+√(5/3) = x2-(5/3) = 0 (3x2−5)=0, is a factor of given polynomial f(x). Now, when we will divide f(x) by (3x2−5) the quotient obtained will also be a factor of f(x) and the remainder will be

0.

Therefore, 3x4 +6x3 −2x2 −10x–5 = (3x2 –5)(x2+2x+1) Now, on further factorizing (x2+2x+1) we get, x2+2x+1 = x2+x+x+1 = 0 x(x+1)+1(x+1) = 0 (x+1)(x+1) = 0 So, its zeroes are given by: x= −1 and x = −1. Therefore, all four zeroes of given polynomial equation are: √(5/3),- √(5/3) , −1 and −1. Hence, is the answer. 4. On dividing x3-3x2+x+2 by a polynomial g(x), the quotient and remainder were x–2 and –2x+4, respectively.

Find g(x). Solution: Given, Dividend, p(x) = x3-3x2+x+2 Quotient = x-2 Remainder = –2x+4 We have to find the value of Divisor, g(x) =? As we know, Dividend = Divisor × Quotient + Remainder ∴ x3-3x2+x+2 = g(x)×(x-2) + (-2x+4) x3-3x2+x+2-(-2x+4) = g(x)×(x-2) Therefore, g(x) × (x-2) = x3-3x2+x+2



Now, for finding g(x) we will divide x3-3x2+x+2 with (x-2)

Therefore, g(x) = (x2–x+1) 5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0 Solutions: According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can

find the value of quotient q(x) and remainder r(x), with the help of below given formula; Dividend = Divisor × Quotient + Remainder ∴ p(x) = g(x)×q(x)+r(x) Where r(x) = 0 or degree of r(x)< degree of g(x). Now let us proof the three given cases as per division algorithm by taking examples for each. (i) deg p(x) = deg q(x) Degree of dividend is equal to degree of quotient, only when the divisor is a constant term. Let us take an example, 3x2+3x+3 is a polynomial to be divided by 3. So, (3x2+3x+3)/3 = x2+x+1 = q(x) Thus, you can see, the degree of quotient is equal to the degree of dividend. Hence, division algorithm is satisfied here. (ii) deg q(x) = deg r(x) Let us take an example , p(x)=x2+x is a polynomial to be divided by g(x)=x. So, (x2+x)/x = x+1 = q(x) Also, remainder, r(x) = 0 Thus, you can see, the degree of quotient is equal to the degree of remainder. Hence, division algorithm is satisfied here. (iii) deg r(x) = 0 The degree of remainder is 0 only when the remainder left after division algorithm is constant. Let us take an example, p(x) = x2+1 is a polynomial to be divided by g(x)=x. So,( x2+1)/x= x=q(x) And r(x)=1



Clearly, the degree of remainder here is 0. Hence, division algorithm is satisfied here.

Exercise 2.4 Page: 36

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2x3+x2-5x+2; -1/2, 1, -2 Solution: Given, p(x) = 2x3+x2-5x+2 And zeroes for p(x) are = 1/2, 1, -2 ∴ p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0 p(1) = 2(1)3+(1)2-5(1)+2 = 0 p(-2) = 2(-2)3+(-2)2-5(-2)+2 = 0 Hence, proved 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2. Now, comparing the given polynomial with general expression, we get; ∴ ax3+bx2+cx+d = 2x3+x2-5x+2 a=2, b=1, c= -5 and d = 2 As we know, if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then; α +β+γ = –b/a αβ+βγ+γα = c/a α βγ = – d/a. Therefore, putting the values of zeroes of the polynomial, α+β+γ = ½+1+(-2) = -1/2 = –b/a αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a α β γ = ½×1×(-2) = -2/2 = -d/a Hence, the relationship between the zeroes and the coefficients are satisfied. (ii) x3-4x2+5x-2 ;2, 1, 1 Solution: Given, p(x) = x3-4x2+5x-2 And zeroes for p(x) are 2,1,1. ∴ p(2)= 23-4(2)2+5(2)-2 = 0 p(1) = 13-(4×12 )+(5×1)-2 = 0 Hence proved, 2, 1, 1 are the zeroes of x3-4x2+5x-2 Now, comparing the given polynomial with general expression, we get; ∴ ax3+bx2+cx+d = x3-4x2+5x-2 a = 1, b = -4, c = 5 and d = -2 As we know, if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then; α + β + γ = –b/a αβ + βγ + γα = c/a α β γ = – d/a. Therefore, putting the values of zeroes of the polynomial, α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a Hence, the relationship between the zeroes and the coefficients are satisfied.



2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Solution: Let us consider the cubic polynomial is ax3+bx2+cx+d and the values of the zeroes of the polynomials be α, β, γ. As per the given question, α+β+γ = -b/a = 2/1 αβ +βγ+γα = c/a = -7/1 α βγ = -d/a = -14/1 Thus, from above three expressions we get the values of coefficient of polynomial. a = 1, b = -2, c = -7, d = 14 Hence, the cubic polynomial is x3-2x2-7x+14 3. If the zeroes of the polynomial x3-3x2+x+1 are a – b, a, a + b, find a and b. Solution: We are given with the polynomial here, p(x) = x3-3x2+x+1 And zeroes are given as a – b, a, a + b Now, comparing the given polynomial with general expression, we get; ∴px3+qx2+rx+s = x3-3x2+x+1 p = 1, q = -3, r = 1 and s = 1 Sum of zeroes = a – b + a + a + b -q/p = 3a Putting the values q and p. -(-3)/1 = 3a a=1 Thus, the zeroes are 1-b, 1, 1+b. Now, product of zeroes = 1(1-b)(1+b) -s/p = 1-b2 -1/1 = 1-b2 b2 = 1+1 = 2 b = √2 Hence,1-√2, 1 ,1+√2 are the zeroes of x3-3x2+x+1. 4. If two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2 ±√3, find other zeroes. Solution: Since this is a polynomial equation of degree 4, hence there will be total 4 roots. Let f(x) = x4-6x3-26x2+138x-35 Since 2 +√3 and 2-√3 are zeroes of given polynomial f(x). ∴ [x−(2+√3)] [x−(2-√3)] = 0 (x−2−√3)(x−2+√3) = 0 On multiplying the above equation we get, x2-4x+1, this is a factor of a given polynomial f(x). Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.



So, x4-6x3-26x2+138x-35 = (x2-4x+1)(x2 –2x−35) Now, on further factorizing (x2–2x−35) we get, x2–(7−5)x −35 = x2– 7x+5x+35 = 0 x(x −7)+5(x−7) = 0 (x+5)(x−7) = 0 So, its zeroes are given by: x= −5 and x = 7. Therefore, all four zeroes of given polynomial equation are: 2+√3 , 2-√3, −5 and 7.

Frequently Asked Questions on Chapter 2- Polynomials

The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x),

in each case? Graphical method to find zeroes:- Total number of zeroes in any polynomial equation = total number of times the

curve intersects x-axis. In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does not cut it at any point.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the

coefficients 4u2+ 8u? 4u(u+2) Therefore, zeroes of polynomial equation 4 + 8u are {0, -2}. Sum of zeroes = 0+(-2) = -2 =-8/4 =(-coefficient

of u)/coefficient of u2 Product of zeroes = 0x-2 = 0 = 0/4 =constant term/coefficient of u2

Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and deg p(x) = deg q(x)?

According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula; Dividend = Divisor × Quotient + Remainder ∴ p(x) = g(x) × q(x) + r(x) Where r(x) = 0 or degree of r(x)< degree of g(x). Now let us proof the three given cases as per division algorithm by taking examples for each. deg p(x) = deg q(x) Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.Let us take an example, 3x2+3x+3 is a polynomial to be divided by 3. So, 3x2+3x+3/3=x2+x+1=q(x) Thus, you can see, the degree of quotient is equal to the degree of dividend. Hence, division algorithm is satisfied here.



Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively?

Let us consider the cubic polynomial is ax3+bx2+cx+d and the values of the zeroes of the polynomials be α, β, γ. As per the given question, α + β + γ = -b/a = 2/1 αβ + βγ + γα = c/a = -7/1 α β γ = -d/a = -14/1 Thus, from above three expressions we get the values of coefficient of polynomial. a = 1, b = -2, c = -7, d = 14 Hence, the cubic polynomial is x3-2x2-7x+14

Find the zeros in the given x^2 – 2x – 8 quadratic polynomial and verify the relationship between the zeroes and

the coefficients. Let us simplify the given quadratic polynomial, x^2 – 2x – 8 = x^2 – 4x + 2x – 8 = x(x – 4) + 2(x – 4) = (x – 4) (x + 2) Therefore, zeros of polynomial equation x^2 – 2x – 8 are (4, -2) So, Sum of zeros = 4 – 2 = 2 = -(-2)/1 = -(Coefficient of x) / (Coefficient of x^2) Product of zeros = 4×(-2) = -8 = -(8)/1 = (Constant term) / (Coefficient of x^2) Find a quadratic polynomial with the given numbers 1/4 , -1 as the sum and product of its zeros respectively. By using the formulas, Sum of zeros = α + β Product of zeros = α β So, Sum of zeros = α + β = 1/4 Product of zeros = α β = -1 Therefore, If α and β are zeros of any quadratic polynomial, then the quadratic polynomial equation can be written

directly as: x^2 – (α + β)x + αβ = 0 x^2 – (1/4)x + (-1) = 0 4x^2 – x – 4 = 0 Hence, 4x^2 – x – 4 is the quadratic polynomial. Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and the product of its

zeros as 2, –7, –14 respectively. Let us consider the cubic polynomial is ax^3 + bx^2 + cx + d and the values of the zeroes of the polynomials are α, β,

γ. So according to the question, α + β + γ = -b/a = 2/1 αβ + βγ + γα = c/a = -7/1 αβγ = -d/a = -14/1 Thus, from the above three expressions we get the values of coefficient of polynomial. a = 1, b = -2, c = -7, d = 14 Hence, the cubic polynomial is x^3 – 2x^2 – 7x + 14

Class 10th Chapter-2 Polynomial Revision Notes & Solution

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