Chapter 6 Factorisation of Algebraic Expressions RD Sharma Solutions Exercise 6.3 Class 9 Maths
![Chapter 6 Factorisation of Algebraic Expressions RD Sharma Solutions Exercise 6.3 Class 9 Maths Chapter 6 Factorisation of Algebraic Expressions RD Sharma Solutions Exercise 6.3 Class 9 Maths](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhBgj6YKOI1BDIB2284tcqYkTmfZxXZ_YPglwCjv1_MZbKEtIL8M6oG-DZ5nqqlcY7BWOqt2_J4Nm0R1opNFWL6xuCb5RRzB7R7f8wv7G29yCzhM8p4UahBGkDQelw8bt1Y4cn9kCf0vVRCVegfhyZwhBJMsucznQwVc2Pr0AkEEv3t15OLoFftUO7gYw/s16000/rd-sharma-solutions-for-class-9-maths-chapter6-factorisation-of-polynomials-exercise-6-3.jpg)
Chapter Name | RD Sharma Chapter 6 Factorisation of Polynomials Exercise 6.3 |
Book Name | RD Sharma Mathematics for Class 10 |
Other Exercises |
|
Related Study | NCERT Solutions for Class 10 Maths |
Exercise 6.3 Solutions
In each of the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the result by actual division : (1 - 8)
1. f(x) = x3 + 4x2 - 3x + 10, g(x) = x + 4
Solution
We have f(x) = x3 + 4x2 - 3x + 10, g(x) = x + 4
Therefore, by remainder theorem when f(x) is divided by g(x) = x - (- 4), the remainder is equal to f(-4)
2. f(x) = 4x4 - 3x3 - 2x2 + x - 7, g(x) = x - 1
Solution
We have,
f(x) = 4x4 - 3x3 - 2x2 + x - 7, and g(x) = x - 1
Therefore by remainder theorem when f(x) is divide by g(x) = x - 1, the remainder is equal to f(+1)
Solution
We have,
f(x) = 2x4 - 6x3 + 2x2 - x + 2, and g(x) = x + 2
Therefore, by remainder theorem when f(x) is divide by g(x) = x - (-2), the remainder is equal to f(-2)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEglNrlIwVuMfpqVQkpjraUBHuXr5VtXucnpjI287ApE55NG2azXBFtj7PrfnewdBvLOOCqQKYQZYfFheCykvto5ll7q8q3jIuzS3WFf0yVoylXdTCHZXp69Max0T4HbVhzSmCXibCv5STwB74RY4Nzc3sTSD_-Hjbt70slEG93MksGWdXWN6MVI89NW/w323-h141/Class%209%20Chapter%20-%206%20Factorization%20of%20Polynomials%20Exercise%20-%206.3%20img%203.JPG)
4. f(x) = 4x3 - 12x3 + 14x - 3, g(x) = 2x - 1
Solution
We have
f(x) = 4x3 - 12x3 + 14x - 3, and g(x) = 2x - 1
Therefore, by remainder theorem when f(x) is divide by g(x) = 2[x - 1/2], the remainder is equal to f(1/2)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9Khvtgp62m5v7dhcfVAIa16SYNrOLZXYGJO9pXpfSiNxjQ2WIyQAxg7dkLOKbJ95GYnNi9gjG2qL9XVuc2Lz6ijm32YSzmzz2Q6aPio5ulQ0jcnVLynxMxhEq2JD-Lq5L2OAodgjqFh11MwxMjjQQ6ZOkDw5Qf3GUXUexCly6GN-iRXekq5KXVwV8/w266-h225/Class%209%20Chapter%20-%206%20Factorization%20of%20Polynomials%20Exercise%20-%206.3%20img%204.JPG)
f(x) = 4x3 - 12x3 + 14x - 3, and g(x) = 2x - 1
Therefore, by remainder theorem when f(x) is divide by g(x) = 2[x - 1/2], the remainder is equal to f(1/2)
5. f(x) = x3 - 6x3 + 2x - 4, g(x) = 1 - 2x
Solution
We have
f(x) = x3 - 6x3 + 2x - 4, and g(x) = 1 - 2x
Therefore, by remainder theorem when f(x) is divided by g(x) = -2(x - 1/2), the remainder is equal to f(1/2)
f(x) = x3 - 6x3 + 2x - 4, and g(x) = 1 - 2x
Therefore, by remainder theorem when f(x) is divided by g(x) = -2(x - 1/2), the remainder is equal to f(1/2)
6. f(x) = x3 - 3x3 + 4, g(x) = x - 2
Solution
We have
f(x) = x3 - 3x3 + 4, and g(x) = x - 2
Therefore, by remainder theorem when f(x) is divided by g(x) = x - 2, the remainder is equal to f(2)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhMNmrUJBnbhpcKaLSVdosAbrO3MQV4FrShO-laswvRxNYVu3YOBojX-FM-mY30gCAU_RXdkPoTGL7WP_1Gp8y5yvMnnnV6aqMmk2wsG0Y-TiVZdZRGu2mdVtptECjOmOhAV4JsIpgC4uOOHt0NHloUOXdCI79OuAxjLHvA2O8Irh1O8MH2PXHdyrOr/s1600/Class%209%20Chapter%20-%206%20Factorization%20of%20Polynomials%20Exercise%20-%206.3%20img%206.JPG)
7. f(x) = 9x3 - 3x3 + x - 5, g(x) = x - 2/3
Solution
We have
f(x) = 9x3 - 3x3 + x - 5, and g(x) = x - 2/3
Therefore, by remainder theorem when f(x) is divided by g(x) = x - 2/3, the remainder is equal to f(2/3)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjQqEbev4e0-uMnJ8m4afuAhAMKElFhaP8L9ARndFADOWaZyNM0QRDLyQxCj9X6cA8URzY2PJR0SjZRduhjgyWPdCmX9mQmN8BsmTS2QWFVQaz_24kZhyfEnGeR4mJXCQvF13xgMctzHswMtLpuXafaahqM1ZBQPeCl8mia0SdVUn1NiPsx5HaxlP3-/s1600/Class%209%20Chapter%20-%206%20Factorization%20of%20Polynomials%20Exercise%20-%206.3%20img%207.JPG)
f(x) = 9x3 - 3x3 + x - 5, and g(x) = x - 2/3
Therefore, by remainder theorem when f(x) is divided by g(x) = x - 2/3, the remainder is equal to f(2/3)
8. f(x) = 3x3 + 2x3 - x3/3 - x/9 + 2/27, g(x) = x+ 2/3
Solution
We have,
9. If the polynomials 2x3 + ax2 + 3x - 5 and x3 + x2 -4x + a leave the same remainder when divided by x - 2, find the value of a.
Solution
Let p(x) = 2x3 + ax2 + 3x - 5 and q(x) = x3 + x2 - 4x + a be the given polynomials
The remainders when p(x) and q(x) are divided by (x - 2) are p(2) and q(2) respectively.
The remainders when p(x) and q(x) are divided by (x - 2) are p(2) and q(2) respectively.
By the given condition we have
10. The polynomials ax3 + 3x2 - 3 and 2x3 - 5x + a when divided by (x - 4) leave the remainder R1 and R2 respectively. Find the values of a in each of the following cases, if
(i) R1 = R2
(ii) R1 + R2 = 0
(iii) 2R1 + R2 = 0
Solution
11. If the polynomials ax3 + 3x2 - 13 and 2x3 - 5x + a, when divided by (x - 2) leave the same remainder, find the value of a.
Solution