ICSE Class 10 Maths Selina Solutions Chapter 8:
ICSE Class 10 Maths Selina Solutions for Chapter 8 Remainder and Factor Theorems provide a detailed guide to understanding and applying these crucial algebraic concepts.
With clear step-by-step explanations and examples these solutions simplify complex problems, making it easier for students to solve them accurately.
By practicing these solutions students can enhance their problem-solving skills and build a strong foundation in algebra which is important for their academic success in mathematics.
ICSE Class 10 Maths Selina Solutions Chapter 8 Remainder and Factor Theorems Overview
These solutions are prepared by subject experts from Physics Wallah for ICSE Class 10 Maths Selina Solutions Chapter 8 Remainder and Factor Theorems. They give clear and detailed explanations, showing step-by-step how to solve problems about remainders and factors of polynomial expressions.
This helps students learn these important math concepts well. By using these solutions, students can improve their problem-solving skills and gain confidence in solving algebra problems.
ICSE Class 10 Maths Selina Solutions Chapter 8 PDF
The PDF link for the ICSE Class 10 Maths Selina Solutions Chapter 8 Remainder and Factor Theorems is available below.
By accessing this PDF, students can enhance their learning experience and effectively prepare for their exams.
ICSE Class 10 Maths Selina Solutions Chapter 8 PDF
ICSE Class 10 Maths Selina Solutions Chapter 8 Remainder and Factor Theorems
Below we have provided ICSE Class 10 Maths Selina Solutions Chapter 8 for the ease of the students –
1. Show that (x – 1) is a factor of x
3
– 7x
2
+ 14x – 8. Hence, completely factorise the given expression.
Solution:
Let f(x) = x
3
– 7x
2
+ 14x – 8
Then, for x = 1
f(1) = (1)
3
– 7(1)
2
+ 14(1) – 8 = 1 – 7 + 14 – 8 = 0
Thus, (x – 1) is a factor of f(x).
Now, performing long division we have
Hence, f(x) = (x – 1) (x
2
– 6x + 8)
= (x – 1) (x
2
– 4x – 2x + 8)
= (x – 1) [x(x – 4) -2(x – 4)]
= (x – 1) (x – 4) (x – 2)
2. Using Remainder Theorem, factorise:
x
3
+ 10x
2
– 37x + 26 completely.
Solution:
Let f(x) = x
3
+ 10x
2
– 37x + 26
From remainder theorem, we know that
For x = 1, the value of f(x) is the remainder
f(1) = (1)
3
+ 10(1)
2
– 37(1) + 26 = 1 + 10 – 37 + 26 = 0
As f(1) = 0, x – 1 is a factor of f(x).
Now, performing long division we have
Thus, f(x) = (x – 1) (x
2
+ 11x – 26)
= (x – 1) (x
2
+ 13x – 2x – 26)
= (x – 1) [x(x + 13) – 2(x + 13)]
= (x – 1) (x + 13) (x – 2)
3. When x
3
+ 3x
2
– mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.
Solution:
Let f(x) = x
3
+ 3x
2
– mx + 4
From the question, we have
f(2) = m + 3
(2)
3
+ 3(2)
2
– m(2) + 4 = m + 3
8 + 12 – 2m + 4 = m + 3
24 – 3 = m + 2m
3m = 21
Thus, m = 7
4. What should be subtracted from 3x
3
– 8x
2
+ 4x – 3, so that the resulting expression has x + 2 as a factor?
Solution:
Let’s assume the required number to be k.
And let f(x) = 3x
3
– 8x
2
+ 4x – 3 – k
From the question, we have
f(-2) = 0
3(-2)
3
– 8(-2)
2
+ 4(-2) – 3 – k = 0
-24 – 32 – 8 – 3 – k = 0
-67 – k = 0
k = -67
Therefore, the required number is -67.
5. If (x + 1) and (x – 2) are factors of x
3
+ (a + 1)x
2
– (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.
Solution:
Let’s take f(x) = x
3
+ (a + 1)x
2
– (b – 2)x – 6
As, (x + 1) is a factor of f(x).
Then, remainder = f(-1) = 0
(-1)
3
+ (a + 1)(-1)
2
– (b – 2) (-1) – 6 = 0
-1 + (a + 1) + (b – 2) – 6 = 0
a + b – 8 = 0 … (1)
And as, (x – 2) is a factor of f(x).
Then, remainder = f(2) = 0
(2)
3
+ (a + 1) (2)
2
– (b – 2) (2) – 6 = 0
8 + 4a + 4 – 2b + 4 – 6 = 0
4a – 2b + 10 = 0
2a – b + 5 = 0 … (2)
Adding (1) and (2), we get
3a – 3 = 0
Thus, a = 1
Substituting the value of a in (i), we get,
1 + b – 8 = 0
Thus, b = 7
Hence, f(x) = x
3
+ 2x
2
– 5x – 6
Now as (x + 1) and (x – 2) are factors of f(x).
So, (x + 1) (x – 2) = x
2
– x – 2 is also a factor of f(x).
Therefore, f(x) = x
3
+ 2x
2
– 5x – 6 = (x + 1) (x – 2) (x + 3)
6. If x – 2 is a factor of x
2
+ ax + b and a + b = 1, find the values of a and b.
Solution:
Let f(x) = x
2
+ ax + b
Given, (x – 2) is a factor of f(x).
Then, remainder = f(2) = 0
(2)
2
+ a(2) + b = 0
4 + 2a + b = 0
2a + b = -4 … (1)
And also given that,
a + b = 1 … (2)
Subtracting (2) from (1), we have
a = -5
On substituting the value of a in (2), we have
b = 1 – (-5) = 6
7. Factorise x
3
+ 6x
2
+ 11x + 6 completely using factor theorem.
Solution:
Let f(x) = x
3
+ 6x
2
+ 11x + 6
For x = -1, the value of f(x) is
f(-1) = (-1)
3
+ 6(-1)
2
+ 11(-1) + 6
= -1 + 6 – 11 + 6 = 12 – 12 = 0
Thus, (x + 1) is a factor of f(x).
Therefore, f(x) = (x + 1) (x
2
+ 5x + 6)
= (x + 1) (x
2
+ 3x + 2x + 6)
= (x + 1) [x(x + 3) + 2(x + 3)]
= (x + 1) (x + 3) (x + 2)
8. Find the value of ‘m’, if mx
3
+ 2x
2
– 3 and x
2
– mx + 4 leave the same remainder when each is divided by x – 2.
Solution:
Let f(x) = mx
3
+ 2x
2
– 3 and g(x) = x
2
– mx + 4
From the question, it’s given that f(x) and g(x) leave the same remainder when divided by (x – 2). So, we have:
f(2) = g(2)
m(2)
3
+ 2(2)
2
– 3 = (2)
2
– m(2) + 4
8m + 8 – 3 = 4 – 2m + 4
10m = 3
Thus, m = 3/10
Benefits of ICSE Class 10 Maths Selina Solutions Chapter 8 Remainder and Factor Theorems
-
Clear Understanding:
These solutions help students grasp the concepts of Remainder and Factor Theorems in a simplified manner making it easier to understand and apply these theorems to various problems.
-
Step-by-Step Solutions:
The solutions are presented in a step-by-step format guiding students through each problem methodically. This approach helps in building a strong foundation in solving polynomial equations.
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Exam Preparation:
With detailed explanations and numerous solved examples, these solutions serve as an excellent resource for exam preparation ensuring students are well-prepared for their tests.
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Practice Questions:
The chapter includes a variety of practice questions that cover different difficulty levels. This allows students to test their understanding and improve their problem-solving skills.
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Confidence Building:
By working through these solutions students can build confidence in their mathematical abilities helping them tackle more complex problems with ease.
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Expert Guidance:
Prepared by subject experts from Physics Wallah, these solutions provide reliable and accurate methods for solving problems, ensuring that students learn the correct techniques.
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Time Management:
Familiarity with the types of questions and the methods to solve them helps students manage their time efficiently during exams.
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