7 - Remainder Theorem

In this post, you can check the solutions to the exercises of Remainder Theorem of Algebra Unit of Readmore Optional Mathematics by DR Simkhada. It consists of complete answers to all odd questions mentioned in the text book.

Here are a few things that you need to consider before checking the solutions:
  1. Remainder theorem states, "If f(x) be a polynomial of degree n where n>0 and g(x) = (x -a) be a binomial such that a is a real number, then the remainder when f(x) is divided by g(x) is given by f(a).
  2. When we have (px - q), a = q/p.
Now, you can check the solutions below. But, make sure to check the description after the solution to know more about this textbook series, author, and the publisher.


1. Using remainder theorem, find the remainder if f(x) is divided by g(x) when,

a) f(x) = x² + 3x + 12 and g(x) = x + 1
Solution:
Comparing (x +1) with (x - a), we get, a = -1
According to remainder theorem, remainder is f(a)
So, f(-1) = (-1)² + 3(-1) + 12
= 1 -3 +12
= 10

c) f(x) = 2x⁵ - 3x² - x + 5 and g(x) = x - 4
Solution:
Comparing (x -4) with (x -a), we get, a = 4
According to remainder theorem, remainder is f(a)
So, f(4) = 2(4)⁵ - 3(4)² - 4 + 5
= 2048 - 48 - 4 + 5
= 2001

e) f(x) = x³ - kx² - 2x + k + 4 and g(x) = x - k
Solution:
Comparing (x-k) with (x -a), we get, a = k
According to remainder theorem, remainder is f(a)
So, f(k) = k³ - k(k)² - 2(k) + k + 4
= k³ - k³ - 2k + k + 4
= 4 - k

g) f(x) = 2x³ - 11x² + 19x - 10 and g(x) = 2x - 5
Solution:
Comparing (2x -5) with (px - q), we get, q/p = 5/2
According to remainder theorem, remainder is f(p/q)
So, f(5/2) = 2(5/2)³ - 11(5/2)² + 19(5/2) - 10
= 125/4 - 275/4 + 95/2 - 10
= (125 - 275 + 190 - 40)/4
= 0/4
= 0

2. Let f(y) = y³ - ay² -2y +1. Find the remainder using synthetic division when f(y) is divided by (y - a). Also, find f(a). Do you see any relation between the remainder and f(a).
Solution:
Here,
f(y) = y³ - ay² -2y + 1
Binomial is (y - a)
Using synthetic division,
a | 1 -a -2   1
   |     a   0 -2a
   |_______________
     1  0  -2  1 -2a
Here, remainder = 1 - 2a
Again,
f(a) = (a)³ - a(a)² - 2a + 1
= a³ - a³ - 2a + 1
= 1 - 2a
Conclusion: Yes, there is a relation between the remainder and f(a). It is that both of them are equal i.e. (1 - 2a).


3. f(x) is the polynomial 4x³ - 12x² + 11x - 3. Use the remainder theorem to find the remainder when f(x) is divided by:

a) (x - 1/2)
Solution:
Comparing (x - 1/2) with (x - a), we get, a = ½
According to remainder theorem, remainder is f(a)
f(1/2) = 4(1/2)³ - 12(1/2)² + 11(1/2) - 3
= 4(1/8) - 12(1/4) + 11/2 - 3
= 1/2 - 3 + 11/2 - 3
= (1+11)/2 - 6
= 12/2 - 6
= 6 - 6
= 0
c) (x - 3/2)
Solution:
Comparing (x - 3/2) with (x - a), we get, a = 3/2
According to remainder theorem, remainder is f(a)
f(3/2) = 4(3/2)³ - 12(3/2)² + 11(3/2) - 3
= 4(27/8) - 12(9/4) + 33/2 - 3
= 27/2 - 3×9 + 33/2 - 3
= (27 + 33)/2 - 27 - 3
= 30 - 30
= 0

4a) If the polynomial 2x³ + 3x² + kx + 4 leaves a remainder 2(4 - k) when divided by (x -2), find the value of k.
Solution:
Comparing (x -2) with (x -a), we get, a = 2
According to remainder theorem, remainder is f(a)
Given,
f(2) = 2(4 - k)
or, 2(4 - k) = f(2)
or, 2(4 - k) = 2(2)³ + 3(2)² + k(2) + 4
or, 8 - 2k = 16 + 12 + 2k + 4
or, 8 - 2k = 32 + 2k
or, 8 - 32 = 2k + 2k
or, - 24 = 4k
or, 4k = -24
So, k = -6


4c) If both ax³ +2x² - 3 and x² -ax +4 leaves the same remainder when divided by (x -2), find a.
Solution:
Let f(x) = ax³ + 2x² - 3
Let g(x) = x² - ax + 4
Comparing (x -2) with (x -a), we get, a = 2
According to remainder theorem and question,
f(a) = g(a)
or, f(2) = g(2)
or, a(2)³ + 2(2)² - 3 = (2)² - a(2) + 4
or, 8a + 8 - 3 = 4 - 2a + 4
or, 8a + 5 = 8 - 2a
or, 8a + 2a = 8 - 5
or, 10a = 3
So, a = 3/10


4e) A polynomial 4x⁴ - (p+1)x² + 8x +14 when divided by 2x+1 leaves the remainder 8.5. Find the value of p by using remainder theorem.
Solution:
Let f(x) = 4x⁴ - (p+1)x² + 8x + 14
Comparing (2x +1) with (px -q), we get, q/p = -1/2
According to the remainder theorem and question, we have,
f(q/p) = 8.5
or, f(-1/2) = 8.5
or, 4(-1/2)⁴ - (p+1)(-1/2)² + 8(-1/2) + 14 = 8.5
or, 4(1/16) - (p+1)(1/4) - 4 + 14 = 8.5
or, 1/4 - (p+1)/4 + 10 = 8.5
or, (1 - p -1)/4 = 8.5 - 10
or, (- p)/4 = -1.5
or, (- p) = (-1.5)×4
or, - p = -6
So, p = 6


5a) The expression ax²+bx+1 has the remainder 2 when it is divided by (x -1) and when it is divided by (x +1), the remainder is 4. Find the values of a and b.
Solution:
Let f(x) = ax² + bx + 1
In condition I,
Comparing (x -1) with (x -a), we get, a = 1
According to remainder theorem and question,
f(a) = 2
or, f(1) = 2
or, a(1)² + b(1) + 1 = 2
or, a + b + 1 = 2
or, a + b = 1
or, a = 1 - b  ----(i)
In condition II,
Comparing (x+1) with (x -a), we get, a = -1
According to remainder theorem and question,
f(a) = 4
or, f(-1) = 4
or, a(-1)² + b(-1) + 1 = 4
or, a - b + 1 = 4
or, a - b = 3
or, a = 3 + b ----(ii)
From equations (i) and (ii), we get,
1 - b = 3 + b
or, 1 - 3 = b + b
or, -2 = 2b
So, b = -1
Put value of b in equation (i), we get,
a = 1 - (-1) = 1 + 1 = 2
Hence, values of a and b are 2 and -1, respectively.


About Readmore Optional Mathematics Book 10

Author: D. R. Simkhada
Editor: I. R. Simkhada

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Remainder Theorem - Solved Exercises | Class 10 Readmore Optional Mathematics is a collection of the solutions related to exercises of Remainder Theorem from the Algebra chapter for Nepal's Secondary Education Examination (SEE) appearing students.