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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
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