CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26
NCERT Solutions for class 10 maths ex 2.3 Polynomials is discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Exercise 2.3 of Class 10 Maths introduces the notion of polynomials in x with real coefficients, the notion of polynomials of degrees 1, 2, 3, quadratic equation, and many more with numerical problems on the Polynomials chapter.
The Class 10th Maths chapter 2 exercise 2.3 moves towards the topics like the class of addition, subtraction and multiplication of polynomials. Also, zeroes/roots of a polynomial function f(x). Also students can find all exercise listed below, practice them to command the maths concepts.
Polynomials Class 10 Maths Chapter 2 Excercise: 2.3
Answer: The polynomial division is carried out as follows
The quotient is x-3 and the remainder is 7x-9
Answer:
The division is carried out as follows
The quotient is $x^2+x-3$
and the remainder is 8
Answer: The polynomial is divided as follows
The quotient is $-x^2-2$ and the remainder is $-5x+10$
$t^2 - 3, 2t^4 + 3t^3 - 2t^2 - 9t - 12$
Answer:
After dividing we got the remainder as zero. So $t^2 - 3$ is a factor of $2t^4 + 3t^3 - 2t^2 - 9t - 12$
$x^2 + 3x + 1, 3x^4 + 5x^3 - 7x^2 + 2x + 2$
Answer: To check whether the first polynomial is a factor of the second polynomial we have to get the remainder as zero after the division
After division, the remainder is zero thus $x^2+3x+1$ is a factor of $3x^4 + 5x^3 - 7x^2 + 2x + 2$
Q2 (3) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: $x^3 - 3x + 1, x^5 - 4x^3 + x^2 + 3x + 1$
Answer: The polynomial division is carried out as follows
The remainder is not zero, there for the first polynomial is not a factor of the second polynomial
Q3 Obtain all other zeroes of $3x^4 + 6x^3 - 2x^2 - 10x - 5$ , if two of its zeroes are $\sqrt {\frac5{3}} \: \:and \: \: - \sqrt {\frac{5}{3}}$
Answer:
Two of the zeroes of the given polynomial are $\sqrt {\frac5{3}} \: \:and \: \: - \sqrt {\frac{5}{3}}$ .
Therefore two of the factors of the given polynomial are $x-\sqrt{\frac{5}{3}}$ and $x+\sqrt{\frac{5}{3}}$
$(x+\sqrt{\frac{5}{3}})\times (x-\sqrt{\frac{5}{3}})=x^{2}-\frac{5}{3}$
$x^{2}-\frac{5}{3}$ is a factor of the given polynomial.
To find the other factors we divide the given polynomial with $3\times (x^{2}-\frac{5}{3})=3x^{2}-5$
The quotient we have obtained after performing the division is $x^{2}+2x+1$
$\\x^{2}+2x+1\\ =x^{2}+x+x+1\\ =x(x+1)+(x+1)\\ =(x+1)^{2}$
(x+1) 2 = 0
x = -1
The other two zeroes of the given polynomial are -1.
Answer: Quotient = x-2
remainder =-2x+4
$g(x)=\frac{x^3-3x^2+3x-2}{x-2}$
Carrying out the polynomial division as follows
$g(x)={x^2-x+1}$
Answer: deg p(x) will be equal to the degree of q(x) if the divisor is a constant. For example
$\\p(x)=2x^2-2x+8\\q(x)=x^2-x+4\\g(x)=2\\r(x)=0$
Q5 (2) Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and deg q(x) = deg r(x)
Answer: Example for a polynomial with deg q(x) = deg r(x) is given below
Answer:
example for the polynomial which satisfies the division algorithm with r(x)=0 is given below
$\\p(x)=x^3+3x^2+3x+5\\q(x)=x^2+2x+1\\g(x)=x+1\\r(x)=4$
The problems from the concepts of division of polynomial p(x) with polynomial g(x) are covered in exercise 2.3 Class 10 Maths. The Introductory numerical questions of NCERT solutions for Class 10 Maths chapter 2 exercise 2.3 is to represent the numerical problems of finding quotient and remainder. And later on questions of Class 10 Maths chapter 2 exercise 2.3 is to find the polynomial if Quotient and remainder is given.
Also Read| Polynomials Class 10 Notes
The solved example before the exercise 2.3 Class 10 Maths and the NCERT solutions for Class 10 Maths chapter 2 exercise 2.3 are important as it covers questions from the basic division of Polynomials
If students can solve each and every question of this exercise 2.3 Class 10 Maths he will be able to grasp the factorization concept of the polynomial as given in Class 10th Maths chapter 2 exercise 2.4
For Class 10 final exams students may get a short answer or long answer questions from the type covered in the Class 10 Maths chapter 2 exercise 2.3
Also see-
Frequently Asked Questions (FAQs)
No, a ≠ 0 because it will turn into a linear equation
When ax² + bx + c =0 has roots/zeros α and β if they are equal, then sign of a (coef of x²) and c (constant term) must be equal
sum of roots = α + β = -(b/a).
Product of roots = α * β = (c/a)
A polynomial with degree = 3 is called Cubic polynomial
Ans Consider the functions, f(x) and g(x) are any two polynomials with for any value of x, g(x) ≠ 0, then we have derived an expression-
f(x) = g(x) × q(x) + r(x)
Dividend = Divisor * Quotient + Remainder
Each term of the first polynomial is multiplied by each term of the second polynomial when the two polynomials are multiplied.
Polynomial '' comes from the word ‘Poly’ (Meaning Many) and ‘nomial’ (in this case meaning Term)-so it means many terms.
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