In the RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions, we are providing the detail about the Factorization of Polynomials with Integral Coefficient using the Factor Theorem. We have given the details based on the CBSE Book of Class 9 and RD Sharma. The Integral Coefficient is the coefficient in an algebraic identity that is an integer. Here learners will get to know the complete details like- Terms, Explanation, Examples, Benefits, frequently asked questions related to the Factorization of Polynomials with Integral Coefficient using the Factor Theorem.
In the attached PDF, there are various types of questions that help students score well in the exam. The solution to each problem is provided with tricks, tips, and shortcuts. Learners can also practice for the exam with these questions by downloading them and practice offline anytime based on their schedule.
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Download RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions PDF
Solutions for Class 9 Maths Chapter 6 Factorization of Polynomials Exercise 6.5
Important Definitions RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions
Suppose ‘P’ is a polynomial with integer coefficients, where P(a) − P(b) is divisible by a−b for any separate integers a and b. In particular, all integer roots of P divide P(0).
Examples of the Factorization of Polynomials with Integral Coefficient using Factor Theorem
Here are some solved examples of the Integral Coefficient using Factor Theorem based on the RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions-
Ques- Use the factor theorem to factorize the following polynomials completely.
x3 – 13x – 12
Given,
f(x)= x3 – 13x – 12
We need to find out at least one root by hit and trial applying the factor theorem to solve this problem.
Here is a trick:
Suppose a polynomial function has integer coefficients. Every rational zero will have the form qp, where ‘p’ is a factor of the constant and ‘q’ is a factor of the leading coefficient.
Here
p= ±1, ±2, ±3, ±4, ±6, ±12 and
q= ±1
Find all combinations of ±qp.
These are the probable roots of the polynomial function.
±1, ±2, ±3, ±4, ±6, ±12
Substituting x=1 in f(x), we get
f(−1)= (–1)3 – 13(–1) – 12
= −1+13−12
= 13+13
=0
Therefore, (x+1) is the factor of f(x)
Now, dividing f(x) by (x+1), we get
x3 – 13x – 12 = (x+1) (x2 – x – 12)
=(x+1) (x2 – 4x + 3x – 12)
=(x+1) {x(x−4) + 3(x−4)}
=(x+1) (x+3) (x−4)
Ques- x3 – 6x2 + 3x + 10
Solution-
x3 – 6x2 + 3x + 10
Solution:
Let p(x) = x3 – 6x2 + 3x + 10
Constant term = 10
Factors of 10 are ±1, ±2, ±5, ±10
Let x + 1 = 0 or x = -1
p(-1) = (-1)3 – 6(-1)2 + 3(-1) + 10 = 10 – 10 = 0
p(-1) = 0
Let x + 2 = 0 or x = -2
p(-2) = (-2)3 – 6(-2)2 + 3(-2) + 10 = -8 – 24 – 6 + 10 = -28
p(-2) ≠ 0
Let x – 2 = 0 or x = 2
p(2) = (2)3 – 6(2)2 + 3(2) + 10 = 8 – 24 + 6 + 10 = 0
p(2) = 0
Let x – 5 = 0 or x = 5
p(5) = (5)3 – 6(5)2 + 3(5) + 10 = 125 – 150 + 15 + 10 = 0
p(5) = 0
Therefore, (x + 1), (x – 2) and (x-5) are factors of p(x)
Hence p(x) = (x + 1) (x – 2) (x-5)
Ques- x4 – 2x3 – 7x2 + 8x + 12
Solution-
p(x) = x4 – 2x3 – 7x2 + 8x + 12
Constant term = 12
Factors of 12 are ±1, ±2, ±3, ±4, ±6, ±12
Let x – 1 = 0 or x = 1
p(1) = (1)4 – 2(1)3 – 7(1)2 + 8(1) + 12 = 1 – 2 – 7 + 8 + 12 = 12
p(1) ≠ 0
Let x + 1 = 0 or x = -1
p(-1) = (-1)4 – 2(-1)3 – 7(-1)2 + 8(-1) + 12 = 1 + 2 – 7 – 8 + 12 = 0
p(-1) = 0
Let x +2 = 0 or x = -2
p(-2) = (-2)4 – 2(-2)3 – 7(-2)2 + 8(-2) + 12 = 16 + 16 – 28 – 16 + 12 = 0
p(-2) = 0
Let x – 2 = 0 or x = 2
p(2) = (2)4 – 2(2)3 – 7(2)2 + 8(2) + 12 = 16 – 16 – 28 + 16 + 12 = 0
p(2) = 0
Let x – 3 = 0 or x = 3
p(3) = (3)4 – 2(3)3 – 7(3)2 + 8(3) + 12 = 0
p(3) = 0
Therefore, (x + 1), (x + 2), (x – 2) and (x-3) are factors of p(x)
Hence p(x) = (x + 1)(x + 2) (x – 2) (x-3)
Frequently Asked Questions (FAQs) of RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions
Ques- How do you factor polynomials with integer coefficients?
Ans- Suppose ‘P’ is a polynomial with integer coefficients, where P(a) − P(b) is divisible by a−b for any separate integers a and b. In particular, all integer roots of P divide P(0).