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NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

EXERCISE 2.3
Ques 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

(ii) p(x) = x4 - 3x2 + 4x + 5, g(x) = x2 + 1 - x
(iii) p(x) = x4 - 5x + 6, g(x) = 2 - x2
Sol: Here, dividend p(x) = x3 - 3x2 + 5x - 3
divisor g(x) = x2 - 2
∴ We have
NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)
Thus, the quotient = (x - 3) and remainder = (7x - 9)

(ii) Here, dividend p(x) = x- 3x2 + 4x + 5
and divisor g(x) = x2 + 1 - x
= x2 - x + 1
∴ We have
NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)
Thus, the quotient is (x2 + x - 3) and remainder = 8

(iii) Here, dividend, p(x) = x4 - 5x + 6
and divisor, g(x) = 2 - x2
= - x2 + 2
∴ We have
NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)
Thus, the quotient = –x2 – 2 and remainder = –5x + 10.

Ques 2: Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)
Sol. (i) Dividing 2t4 + 3t3 - 2t2 - 9t - 12 by t2 - 3, we have:

NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

∵ Remainder = 0
∴ (t2 - 3) is a factor of 2t4 + 3t3 - 2t2 - 9t - 12.

(ii) Dividing 3x4 + 5x3 - 7x2 + 2x + 2 by x2 + 3x + 1, we have:

NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

∵ Remainder = 0.
∴ x2 + 3x + 1 is a factor of 3x4 + 5x3 - 7x+ 2x + 2.

(iii) Dividing x5 - 4x3 + x2 + 3x + 1 by x3 - 3x + 1, we get:

NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)
∴ The remainder = 2, i.e., remainder ≠ 0
∴ x3 - 3x + 1 is not a factor of x- 4x3 + x2 + 3x + 1.

Ques 3: Obtain all other zeroes of 3x4 + 6x3 - 2x2 - 10x - 5, if two of its zeroes NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)are - NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3).
Sol: We have p(x) = 3x4 + 6x- 2x2 - 10x - 5
α =NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3) and β = - NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3) is a factor of p (x).

Now, let us divide 3x4 + 6x3 - 2x2 - 10x - 5 by NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3) .

NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)
NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

 For p (x) = 0, we have

NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

i.e., NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

or 3x + 3 = 0 ⇒ x = - 1
or x + 1 = 0 ⇒ x = - 1
Thus, all the other zeroes of the given polynomial are - 1 and - 1.

Ques 4: On dividing x- 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x - 2 and -2x + 4 respectively. Find g(x).
Sol: Here,
Dividend p(x) = x3 - 3x2 + x + 2
Divisor = g(x)
Quotient = (x - 2)
Remainder = (-2x + 4)
Since,

(Quotient × Divisor) + Remainder = Dividend 

∴ [(x − 2) × g(x)] + [(−2x + 4)] = x3 − 3x2 + x + 2
⇒ (x − 2) × g(x)= x3 − 3x2 + x + 2 − (−2x + 4)
= x3 − 3x2 + x + 2 + 2x −4

= x3 - 3x2 + 3x - 2
∴  NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

Now, dividing x3 - 3x+ 3x - 2 by x - 2, we have
NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

∴  NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

Thus, the required divisor g(x) = x- x + 1.

Ques 5: Give example 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

Sol: 
(i) p(x),  g(x), q(x), r(x)
deg p(x) = deg q(x)
∴ both g(x) and r(x) are constant terms.
p(x) = 2x2- 2x + 14
g(x) = 2
q(x) = x2 - x + 7
r(x) = 0

(ii) deg q(x) = deg r(x)
∴ this is possible when deg of both q(x) and r(x) should be less than p(x) and g(x).
p(x) = x3 + x2 + x + 1
g(x) = x2 - 1
q(x) = x + 1, r(x) = 2x + 2
(iii) deg r(x) is 0.
This is possible when product of q(x) and g(x) form a polynomial whose degree is equal to degree of p(x) and constant term.

NOTE: We have given one example for each of the above cases, however, there can be several examples for them.

The document NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3) is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
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FAQs on NCERT Solutions for Class 8 Maths Chapter 2 - Polynomials (Exercise 2.3)

1. What is a polynomial?
Ans. A polynomial is an algebraic expression consisting of variables, coefficients, and exponents, combined using addition, subtraction, and multiplication operations. It can have one or more terms, and each term can have a different degree.
2. How do you identify the degree of a polynomial?
Ans. To identify the degree of a polynomial, you need to find the highest power of the variable in the polynomial. For example, in the polynomial 3x^2 + 5x - 2, the degree is 2 because the highest power of x is 2.
3. What is the difference between a monomial and a binomial?
Ans. A monomial is a polynomial with only one term, while a binomial is a polynomial with exactly two terms. For example, 3x^2 is a monomial, and 3x^2 + 5x is a binomial.
4. How do you add or subtract polynomials?
Ans. To add or subtract polynomials, you simply combine like terms. Like terms have the same variable(s) raised to the same power(s). Add the coefficients of the like terms while keeping the variable(s) and exponent(s) unchanged.
5. Can a polynomial have a negative degree?
Ans. No, a polynomial cannot have a negative degree. The degree of a polynomial is always a non-negative integer. The degree represents the highest power of the variable in the polynomial, and negative powers are not allowed in polynomials.
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