= 2 + 2 + 8 – 8 = 4 Find the value of x3 + y3 + z3 – 3xyz if x2 + y2 + z2 = 83 and x + y + z = 15, From algebraic identities, we know that (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca), (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + xz), From the question, x2 + y2 + z2 = 83 and x + y + z = 15. We have, p(x) = 3x3+7x. Thus, the value of 5x – 4x2 + 3 at x = 0 is 3. Thus, the required remainder is 5a. So, we write a = 32c, where c is an integer. Answer: They are consistent because they deal with two different situations. Question 1. Class 10 Maths Solutions Chapter 14 (ii) The given polynomial is 4- y2. Solution: (i) 27y3 + 125z3 (iii) Segment: Solution: (v) x10+ y3+t50 Thus, the required remainder = \(\frac { 27 }{ 8 }\). … [Taking square root of both sides] A rational number between two rational numbers a and b Then, x + y + z = 28 – 15 – 13 = 0 [Answer: 4xy + 4zx], Factorise: 125x³ + 8y³ + Z³ – 30xyz. (i) 4x2 – 3x + 7 = x2 + 4y2 + 16z2 + 4xy + 16yz + 8 zx, (ii) (2x – y + z)2 = (2x)2 + (- y)2 + z2 + 2 (2x) (- y)+ 2 (- y) (z) + 2 (z) (2x)

∴ p(1) = (1)2 – 1 = 1 – 1=0 ∴ p(-2) = (-2)3 + 3(-2)2+ 3(-2) + 1 So, it is a quadratic polynomial.

The coefficient of x2 is 0. Hence, if x + y + z = 0, then Also, p(-1) = (-1)2 -1 = 1 – 1 = 0 and BD = BC + CD. (i) Here, p(x) = x2 + x + k a = b + c. (i) parallel lines i. = 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)] ∴ p(o) = (0)2 = 0 Here we have given Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Exercise 2.1.

Since, \(p(\frac { 1 }{ 2 } )\) ≠ 0, so, x = \(\frac { 1 }{ 2 }\) is not a zero of 2x + 1. i. When you understand the formulas behind each Polynomials topics then it would be easier to solve the most complex problems related to a Polynomials too. (i) We have, p (x)= 2x3 + x2 – 2x – 1 and g (x) = x + 1 \], \[ Cubic\; Polynomial = x^3 + 4x^2 + 5\;\;{\rm{etc}}{\rm{.}} k = -2 – √2 = -(2 + √2), (iii) Here, p (x) = kx2 – √2 x + 1 = x3 + y3 + z3 – 3xyz = L.H.S. Polynomials formulas are very helpful for better scores in the exam. (ii) x3 – y3 = (x – y) (x2 + xy + y2) = \(\frac { 1 }{ 2 }\)(x + y + z)[(x – y)2+(y – z)2+(z – x)2] Since, p(x) = 0 If a point C lies between two points A and B such that AC = BC, then prove that \(A C=\frac{1}{2} A B\). = (3 – 5a)3 => 25x2 – 35x + 12 = 5x(5x – 3) – 4(5x – 3). = 10000 + 1000 + 21=11021, (ii) We have, 95 x 96 = (100 – 5) (100 – 4) which is not a whole number.

Find the value of a and b. Show that 4√2 is an irrational number. ‘O’ – Fixed point 2.10, if AC = BD, then prove that AB = CD. the remainder is not 0. (iii) p (x) = kx2 – √2 x + 1 Check the below NCERT MCQ Questions for Class 9 Maths Chapter 2 Polynomials with Answers Pdf free download. [Using a3 + b3 + 3 ab(a + b) = (a + b)3] Here we must know the point and straight line. If you have any query regarding Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Exercise 2.1, drop a comment below and we will get back to you at the earliest. [Answer: x³ – 3x + 1], Give one example of each monomial, binomial and quadratic polynomial. = x2 (x + 1) – 4x(x + 1) – 5(x + 1) We have, (x + 4) (x + 10) = x2+(4 + 10) x + (4 x 10) = 1 – 3 + 3 – 1 + 1 = 1 (i) We have, 99 = (100 -1)

= (100)3 – 13 – 3(100)(1)(100 -1) Hence, verified.

Answer: Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases Class 10 Maths Solutions Chapter 12 (ii) p(x)= x3 + 3x2 + 3x + 1, g (x) = x + 2 0. (ii) 4 – y2 ⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x).

Now, a > b and a > c. = 4k x (3y + 5) x (y – 1) ⇒ 3x – 2 = 0 Question 2. (i) The given polynomial is 5x3 + 4x2 + 7x. Page Contents [ show] = (2a)3 + (b)3 + 3(2a)(b)(2a + b) A – Starting point We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz Solution: Solution: Class 9 Maths Chapter 2 Polynomials NCERT Book PDF Download. (iv) (3a -7b – c)z Factorise each of the following (iv) The zero of x + π is -π. i.e. A number of lines can pass through a single point. Thus, 2y3 + y2 – 2y – 1 Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1. We hope the UP Board Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद) help you. Because there is a unique line that passes through two points. Thus, zero of 2x + 5 is \(\frac { -5 }{ 2 }\) .

Very nice app and there question is very useful to me, there question is very interesting . This chapter talks about; Polynomials in One Variable; Zeroes of a Polynomial; Remainder Theorem; Factorization of Polynomials; Algebraic Identities The expression is x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2-xy-yz-xz), x^2 +y^2 + z^2 = 83, xy+yz+sa = 71 (i) The given polynomial is 2 + x2 + x. = – 5x – 4x2 + 3 = -9 + 3 = -6 Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2). (iii) We have, = x3 – 4x2 + x + 6 and g (x) = x – 3 Class 9 Maths all chapters solutions are also given in PDF. Class 10 Maths Solutions Chapter 13 ⇒ cx + d = 0 ⇒ cx = -d ⇒ \( x =-\frac { d }{ c }\)

Solution: In ∆OAB, m∠OAB = 90° CBSE recommends NCERT books and most of the questions in … So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that We know that, Since, p(x) = 0 (iv) We have y + \(y+\frac { 2 }{ y }\) = y + 2.y-1 = (2x + 3y – 4z)2 = (2x + 3y + 4z) (2x + 3y – 4z), (ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz ∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1

= (y – 1)(2y2 + 2y + y + 1) Since, x + y + z = 0 Solution: [Hint See question 9] What are the possible expressions for the dimensions of the cuboids whose volumes are given below? Question 2.

(vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers. (i) 8a3 +b3 +12a2b+6ab2 If Angle between \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{OB}}\) is 90°, then only lines are perpendicular mutually. Give reasons for your answers. = 1000300 – 30001 = 970299, (ii) We have, 102 =100 + 2 Solution:

These are the most basic and the most common questions for exams.

∴ 4√2 is an irrational number. ⇒ (x – y)(x2 + y2 + xy) = x3 – y3 [Answer: 3], If p(x) = x³ – 3x² + 2x – 3 find the value of p(1) + p(–1).

Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1. Class 10 Maths Solutions Chapter 10, Class 10 Maths Solutions Chapter 11 = 4k[3y2 – 3y + 5y – 5] Ex 2.1 Class 9 Maths Question 2. = -14 + 13 (ii) 4y2-4y + 1 Explain. (ii) (28)3 + (- 15)3 + (- 13)3 p(2) = (2)3 = 8 Solution: Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5 (vii) 7x3 Question and Answer forum for K12 Students. Use suitable identities to find the following products (iv) 64a3 -27b3 -144a2b + 108ab2 E.g 2: If the human body is full, fingers are its parts. P(2) = (2 – 1)(2 + 1) = (1)(3) = 3.

(i) x2+ x We know that The coordinate of the point Q is √10 .

Since, p(1) = (1)2 +1 + k and zero of 7 + 3x is \(-\frac { 7 }{ 3 }\). ∴ (28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13) (i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3 So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that Write the degree of each of the following polynomials. p(1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1 Solution: Let us assume that 4√2 is a rational number . Factorise the following using appropriate identities Show that (x – 5) is a factor of x³ – 3x² – 4x – 30 by Remainder theorem. Polynomials formulas are very helpful for better scores in the exam.

False. Solution: ∴ Our assumption that 3 – √5 is a rational number is wrong. Solution: Since, p(0) = 0, so, x = 0 is a zero of x2. Point: A point is that which has no part. Important questions for Class 9 Maths Chapter 2 Polynomials are provided here to help the CBSE students score well in their Class 9 Maths exam.

⇒ (x + y)[(x + y)2-3xy] = x3 + y3 Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3), (iii) We have, 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 R.H.S x² + y² + z² = 83 (x + a) (x + b) = x2 + (a + b) x + ab Then C and D are not different, they are the same. Save my name, email, and website in this browser for the next time I comment. If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz. Write the coefficients of x2 in each of the following Since, a and b are integers, \(\frac { a }{ b }\) – 3 is a rational

2AC = 2 AD. Let p(x) = x3 + 3x2 + 3x +1 (i) 9x2 + 6xy + y2

To ace in your exam preparation, you can refer to the 9th Class NCERT Solutions prevailing in NCERT e-Book. ∴ (998)3 = (1000-2)3 = 9a2 + 49b2 + c2 – 42ab + 14bc – 6ac, (v)(- 2x + 5y- 3z)2 = (- 2x)2 + (5y)2 + (- 3z)2 + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x) thank you for everything, Please add some more questions. Why we subtracted 83-71?? 3. To Prove : AB = CD

(ii) x – x3 thank you, Nice app but difficult to do the question plz make it.

∴ p(1) = (1)2 – 1 = 1 – 1=0 ∴ p(-2) = (-2)3 + 3(-2)2+ 3(-2) + 1 So, it is a quadratic polynomial.

The coefficient of x2 is 0. Hence, if x + y + z = 0, then Also, p(-1) = (-1)2 -1 = 1 – 1 = 0 and BD = BC + CD. (i) Here, p(x) = x2 + x + k a = b + c. (i) parallel lines i. = 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)] ∴ p(o) = (0)2 = 0 Here we have given Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Exercise 2.1.

Since, \(p(\frac { 1 }{ 2 } )\) ≠ 0, so, x = \(\frac { 1 }{ 2 }\) is not a zero of 2x + 1. i. When you understand the formulas behind each Polynomials topics then it would be easier to solve the most complex problems related to a Polynomials too. (i) We have, p (x)= 2x3 + x2 – 2x – 1 and g (x) = x + 1 \], \[ Cubic\; Polynomial = x^3 + 4x^2 + 5\;\;{\rm{etc}}{\rm{.}} k = -2 – √2 = -(2 + √2), (iii) Here, p (x) = kx2 – √2 x + 1 = x3 + y3 + z3 – 3xyz = L.H.S. Polynomials formulas are very helpful for better scores in the exam. (ii) x3 – y3 = (x – y) (x2 + xy + y2) = \(\frac { 1 }{ 2 }\)(x + y + z)[(x – y)2+(y – z)2+(z – x)2] Since, p(x) = 0 If a point C lies between two points A and B such that AC = BC, then prove that \(A C=\frac{1}{2} A B\). = (3 – 5a)3 => 25x2 – 35x + 12 = 5x(5x – 3) – 4(5x – 3). = 10000 + 1000 + 21=11021, (ii) We have, 95 x 96 = (100 – 5) (100 – 4) which is not a whole number.

Find the value of a and b. Show that 4√2 is an irrational number. ‘O’ – Fixed point 2.10, if AC = BD, then prove that AB = CD. the remainder is not 0. (iii) p (x) = kx2 – √2 x + 1 Check the below NCERT MCQ Questions for Class 9 Maths Chapter 2 Polynomials with Answers Pdf free download. [Using a3 + b3 + 3 ab(a + b) = (a + b)3] Here we must know the point and straight line. If you have any query regarding Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Exercise 2.1, drop a comment below and we will get back to you at the earliest. [Answer: x³ – 3x + 1], Give one example of each monomial, binomial and quadratic polynomial. = x2 (x + 1) – 4x(x + 1) – 5(x + 1) We have, (x + 4) (x + 10) = x2+(4 + 10) x + (4 x 10) = 1 – 3 + 3 – 1 + 1 = 1 (i) We have, 99 = (100 -1)

= (100)3 – 13 – 3(100)(1)(100 -1) Hence, verified.

Answer: Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases Class 10 Maths Solutions Chapter 12 (ii) p(x)= x3 + 3x2 + 3x + 1, g (x) = x + 2 0. (ii) 4 – y2 ⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x).

Now, a > b and a > c. = 4k x (3y + 5) x (y – 1) ⇒ 3x – 2 = 0 Question 2. (i) The given polynomial is 5x3 + 4x2 + 7x. Page Contents [ show] = (2a)3 + (b)3 + 3(2a)(b)(2a + b) A – Starting point We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz Solution: Solution: Class 9 Maths Chapter 2 Polynomials NCERT Book PDF Download. (iv) (3a -7b – c)z Factorise each of the following (iv) The zero of x + π is -π. i.e. A number of lines can pass through a single point. Thus, 2y3 + y2 – 2y – 1 Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1. We hope the UP Board Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद) help you. Because there is a unique line that passes through two points. Thus, zero of 2x + 5 is \(\frac { -5 }{ 2 }\) .

Very nice app and there question is very useful to me, there question is very interesting . This chapter talks about; Polynomials in One Variable; Zeroes of a Polynomial; Remainder Theorem; Factorization of Polynomials; Algebraic Identities The expression is x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2-xy-yz-xz), x^2 +y^2 + z^2 = 83, xy+yz+sa = 71 (i) The given polynomial is 2 + x2 + x. = – 5x – 4x2 + 3 = -9 + 3 = -6 Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2). (iii) We have, = x3 – 4x2 + x + 6 and g (x) = x – 3 Class 9 Maths all chapters solutions are also given in PDF. Class 10 Maths Solutions Chapter 13 ⇒ cx + d = 0 ⇒ cx = -d ⇒ \( x =-\frac { d }{ c }\)

Solution: In ∆OAB, m∠OAB = 90° CBSE recommends NCERT books and most of the questions in … So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that We know that, Since, p(x) = 0 (iv) We have y + \(y+\frac { 2 }{ y }\) = y + 2.y-1 = (2x + 3y – 4z)2 = (2x + 3y + 4z) (2x + 3y – 4z), (ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz ∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1

= (y – 1)(2y2 + 2y + y + 1) Since, x + y + z = 0 Solution: [Hint See question 9] What are the possible expressions for the dimensions of the cuboids whose volumes are given below? Question 2.

(vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers. (i) 8a3 +b3 +12a2b+6ab2 If Angle between \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{OB}}\) is 90°, then only lines are perpendicular mutually. Give reasons for your answers. = 1000300 – 30001 = 970299, (ii) We have, 102 =100 + 2 Solution:

These are the most basic and the most common questions for exams.

∴ 4√2 is an irrational number. ⇒ (x – y)(x2 + y2 + xy) = x3 – y3 [Answer: 3], If p(x) = x³ – 3x² + 2x – 3 find the value of p(1) + p(–1).

Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1. Class 10 Maths Solutions Chapter 10, Class 10 Maths Solutions Chapter 11 = 4k[3y2 – 3y + 5y – 5] Ex 2.1 Class 9 Maths Question 2. = -14 + 13 (ii) 4y2-4y + 1 Explain. (ii) (28)3 + (- 15)3 + (- 13)3 p(2) = (2)3 = 8 Solution: Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5 (vii) 7x3 Question and Answer forum for K12 Students. Use suitable identities to find the following products (iv) 64a3 -27b3 -144a2b + 108ab2 E.g 2: If the human body is full, fingers are its parts. P(2) = (2 – 1)(2 + 1) = (1)(3) = 3.

(i) x2+ x We know that The coordinate of the point Q is √10 .

Since, p(1) = (1)2 +1 + k and zero of 7 + 3x is \(-\frac { 7 }{ 3 }\). ∴ (28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13) (i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3 So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that Write the degree of each of the following polynomials. p(1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1 Solution: Let us assume that 4√2 is a rational number . Factorise the following using appropriate identities Show that (x – 5) is a factor of x³ – 3x² – 4x – 30 by Remainder theorem. Polynomials formulas are very helpful for better scores in the exam.

False. Solution: ∴ Our assumption that 3 – √5 is a rational number is wrong. Solution: Since, p(0) = 0, so, x = 0 is a zero of x2. Point: A point is that which has no part. Important questions for Class 9 Maths Chapter 2 Polynomials are provided here to help the CBSE students score well in their Class 9 Maths exam.

⇒ (x + y)[(x + y)2-3xy] = x3 + y3 Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3), (iii) We have, 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 R.H.S x² + y² + z² = 83 (x + a) (x + b) = x2 + (a + b) x + ab Then C and D are not different, they are the same. Save my name, email, and website in this browser for the next time I comment. If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz. Write the coefficients of x2 in each of the following Since, a and b are integers, \(\frac { a }{ b }\) – 3 is a rational

2AC = 2 AD. Let p(x) = x3 + 3x2 + 3x +1 (i) 9x2 + 6xy + y2

To ace in your exam preparation, you can refer to the 9th Class NCERT Solutions prevailing in NCERT e-Book. ∴ (998)3 = (1000-2)3 = 9a2 + 49b2 + c2 – 42ab + 14bc – 6ac, (v)(- 2x + 5y- 3z)2 = (- 2x)2 + (5y)2 + (- 3z)2 + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x) thank you for everything, Please add some more questions. Why we subtracted 83-71?? 3. To Prove : AB = CD

(ii) x – x3 thank you, Nice app but difficult to do the question plz make it.