Find the best tutors and institutes for Class 10 Tuition
Q5(ii):
Factorise :
(ii) $x^3 – 3x^2 – 9x – 5$
Solution :
Initial Setup & Given Polynomial
We are tasked with factorising the cubic polynomial. Let the given polynomial be defined as:
$P(x) = x^3 - 3x^2 - 9x - 5$
Step 1: Applying the Rational Root Theorem & Factor Theorem
To find the first linear factor of the cubic polynomial, we utilize the Rational Root Theorem. The theorem states that any rational root of the polynomial must be a divisor of the constant term ($-5$) divided by a divisor of the leading coefficient ($1$).
The possible rational roots are the divisors of $-5$, which are: $\pm 1, \pm 5$.
We evaluate $P(x)$ at these test values to find a root (where $P(x) = 0$):
- Testing $x = 1$:
$P(1) = (1)^3 - 3(1)^2 - 9(1) - 5 = 1 - 3 - 9 - 5 = -16$
Since $P(1) \neq 0$, $x = 1$ is not a root. - Testing $x = -1$:
$P(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5$
$P(-1) = -1 - 3(1) + 9 - 5$
$P(-1) = -1 - 3 + 9 - 5 = 0$
Since $P(-1) = 0$, we conclude [By the Factor Theorem] that $(x + 1)$ is a factor of the polynomial $P(x)$.
Step 2: Algebraic Manipulation (Factorisation by Grouping)
Knowing that $(x + 1)$ is a factor, we can rewrite the terms of the original polynomial $P(x)$ such that $(x + 1)$ can be factored out from grouped terms. This method is mathematically more elegant than polynomial long division.
We systematically split the middle terms to force the appearance of $(x + 1)$:
$P(x) = x^3 - 3x^2 - 9x - 5$
[Split $-3x^2$ into $+x^2 - 4x^2$ to pair with $x^3$]
$P(x) = x^3 + x^2 - 4x^2 - 9x - 5$
[Split $-9x$ into $-4x - 5x$ to pair with $-4x^2$]
$P(x) = x^3 + x^2 - 4x^2 - 4x - 5x - 5$
[Group the terms into pairs]
$P(x) = (x^3 + x^2) - (4x^2 + 4x) - (5x + 5)$
[Factor out the greatest common monomial from each group]
$P(x) = x^2(x + 1) - 4x(x + 1) - 5(x + 1)$
[Factor out the common binomial $(x + 1)$]
$P(x) = (x + 1)(x^2 - 4x - 5)$
Step 3: Factorising the Quadratic Quotient
We are now left with a linear factor and a quadratic factor: $Q(x) = x^2 - 4x - 5$. We must factorise this quadratic expression by splitting the middle term.
We seek two numbers that multiply to the constant term ($-5$) and add up to the coefficient of the middle term ($-4$). These numbers are $-5$ and $1$.
$x^2 - 4x - 5$
$= x^2 - 5x + 1x - 5$
$= x(x - 5) + 1(x - 5)$
$= (x - 5)(x + 1)$
Step 4: Final Synthesis of Factors
Substitute the fully factorised quadratic back into the equation from Step 2:
$P(x) = (x + 1) \cdot [(x - 5)(x + 1)]$
$P(x) = (x + 1)(x + 1)(x - 5)$
$P(x) = (x + 1)^2(x - 5)$
Visual Verification: Graph of the Polynomial
The graph below illustrates the function $y = x^3 - 3x^2 - 9x - 5$. Notice that the curve is tangent to the x-axis at $x = -1$ (indicating a root with a multiplicity of 2, hence $(x+1)^2$) and intersects the x-axis at $x = 5$ (indicating a root with a multiplicity of 1, hence $(x-5)$).
Final Solution: The completely factorised form of the polynomial $x^3 - 3x^2 - 9x - 5$ is $(x + 1)^2(x - 5)$.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.3
- Q1(i): Determine which of the following polynomials has $(x + 1)$ a factor : (i) $x^3 + x^2 + x + 1$
- Q1(ii): Determine which of the following polynomials has $(x + 1)$ a factor : (ii) $x^4 + x^3 + x^2 + x + 1$
- Q1(iii): Determine which of the following polynomials has $(x + 1)$ a factor : (iii) $x^4 + 3x^3 + 3x^2 + x + 1$
- Q1(iv): Determine which of the following polynomials has $(x + 1)$ a factor : (iv) $x^3 – x^2 – (2 + \sqrt{2})x + \sqrt{2}$
- Q2(i): Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: (i) $p(x) = 2x^3 + x^2 – 2x – 1, g(x) = x + 1$
- Q2(ii): Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: (ii) $p(x) = x^3 + 3x^2 + 3x + 1, g(x) = x + 2$
- Q2(iii): Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: (iii) $p(x) = x^3 – 4x^2 + x + 6, g(x) = x – 3$
- Q3(i): Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases: (i) $p(x) = x^2 + x + k$
- Q3(ii): Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases: (ii) $p(x) = 2x^2 + kx + \sqrt{2}$
- Q3(iii): Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases: (iii) $p(x) = kx^2 – \sqrt{2}x + 1$
- Q3(iv): Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases: (iv) $p(x) = kx^2 – 3x + k$
- Q4(i): Factorise : (i) $12x^2 – 7x + 1$
- Q4(ii): Factorise : (ii) $2x^2 + 7x + 3$
- Q4(iii): Factorise : (iii) $6x^2 + 5x – 6$
- Q4(iv): Factorise : (iv) $3x^2 – x – 4$
- Q5(i): Factorise : (i) $x^3 – 2x^2 – x + 2$
- Q5(iii): Factorise : (iii) $x^3 + 13x^2 + 32x + 20$
- Q5(iv): Factorise : (iv) $2y^3 + y^2 – 2y – 1$
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
Top Tutors who teach Polynomials
Super Tutor
Basaveshwara Nagar, Bangalore
12 yrs of
Exp
14 students
3975 hrs of classes completed
800
per hour
Online Classes
With over 15 years of dedicated teaching experience, I am an accomplished and qualified educator specializing in Spoken English, Math, Science, Social Science, and Kannada Language. My expertise extends across various educational boards, including CBSE, ICSE, and state boards. Passionate about unraveling the mysteries of mathematical problems and chemical equations. I have garnered recognition with the prestigious Best Teacher Award for orchestrating state-level Science exams. Having contributed my skills to the esteemed MaxMuller Public School in Bangalore, I am committed to fostering a nurturing and inspiring learning environment. Join me on this educational journey where knowledge meets enthusiasm!
For the last one and a half months, it has been going well; it is the first review in the initial stage. Again, I will review as a class exam, the results come up, but the teaching to date has been very good.
Top Tutor
Gharoli Gharoli, Delhi
17 yrs of
Exp
2 students
204 hrs of classes completed
400
per hour
Online Classes
I bring with me 17 years of diverse experience in the pharmaceutical industry along with academic expertise, combining strong practical knowledge with a passion for teaching. I am dedicated to simplifying complex concepts and delivering them in an engaging manner, ensuring that students not only gain subject knowledge but also develop the skills, confidence, and potential required to achieve academic excellence and become industry-ready professionals.
Very friendly teacher and explain very well and explain in detail overall teacher is very good and friendly.
Top Tutor
Mira Road East, Mumbai
20 yrs of
Exp
11 students
350
per hour
Online Classes
Student's Home
Tutor's home
25 years of private teaching experience in maths and science at Mumbai western line.
Top Tutor
Burul Rd Burul, Burul
5 yrs of
Exp
100
per hour
Online Classes
Student's Home
Tutor's home
I am a teacher of math. I am giving home/online/tutor home tuition since 2016. My key skills are computer language. I am b. Tech graduate. I can teach short tricks that can help students in their future life.
Top Tutor
Kharodi West, Mumbai
10 yrs of
Exp
1 student
700
per hour
Student's Home
An accommodating and versatile individual with the talent to develop inspiring hands-on lessons that will capture a child's imagination and breed success.
Find more Tutor for Polynomials in your City
- Bangalore Mathematics Tutors
- Delhi Mathematics Tutors
- Chennai Mathematics Tutors
- Gurgaon Mathematics Tutors
- Noida Mathematics Tutors
- Hyderabad Mathematics Tutors
- Mumbai Mathematics Tutors
- Chandigarh Mathematics Tutors
- Pune Mathematics Tutors
- Ghaziabad Mathematics Tutors
- Jaipur Mathematics Tutors
- Surat Mathematics Tutors
