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Q5(iv):
Factorise :
(iv) $2y^3 + y^2 – 2y – 1$
Solution :
Initial Setup & Given Polynomial
We are tasked with factorising the following cubic polynomial:
$P(y) = 2y^3 + y^2 - 2y - 1$
Because this is a polynomial of degree 3, it can have up to three linear factors. We will approach this factorization using the Method of Grouping, which is highly efficient for four-term polynomials where a proportional relationship exists between the coefficients.
Step 1: Grouping the Terms
We begin by dividing the four terms of the polynomial into two distinct pairs. This allows us to extract the Greatest Common Factor (GCF) from each pair independently.
$P(y) = (2y^3 + y^2) - (2y + 1)$
[Per the Associative Property of Addition, grouping terms does not alter the value of the expression. Note that factoring out a negative sign from the last two terms changes $-2y - 1$ to $-(2y + 1)$].
Step 2: Extracting the Greatest Common Factor (GCF)
Next, we factor out the GCF from each binomial group.
- From the first group $(2y^3 + y^2)$, the GCF is $y^2$. Factoring this out yields: $y^2(2y + 1)$.
- From the second group $-(2y + 1)$, the GCF is $1$. Factoring this out yields: $-1(2y + 1)$.
Substituting these back into the expression, we obtain:
$P(y) = y^2(2y + 1) - 1(2y + 1)$
Step 3: Factoring out the Common Binomial
Observe that the binomial $(2y + 1)$ is now a common factor in both terms. We can factor this binomial out of the entire expression.
$P(y) = (2y + 1)(y^2 - 1)$
[Per the Distributive Property of Multiplication over Addition, $ab + ac = a(b+c)$, where $a = (2y + 1)$].
Step 4: Applying the Difference of Squares Identity
The expression is now a product of a linear binomial and a quadratic binomial. The quadratic factor, $(y^2 - 1)$, is a classic "Difference of Squares."
[Per the fundamental algebraic identity: $a^2 - b^2 = (a - b)(a + b)$]
Letting $a = y$ and $b = 1$, we expand the quadratic term:
$y^2 - 1 = (y - 1)(y + 1)$
Substituting this back into our main equation yields the complete prime factorization of the cubic polynomial:
$P(y) = (2y + 1)(y - 1)(y + 1)$
Verification via the Factor Theorem (Analytical Check)
To ensure absolute mathematical rigor, we can verify our factors using the Factor Theorem, which states that a polynomial $P(y)$ has a factor $(y - c)$ if and only if $P(c) = 0$.
Let us test the root $y = 1$ (derived from the factor $y - 1$):
$P(1) = 2(1)^3 + (1)^2 - 2(1) - 1$
$P(1) = 2(1) + 1 - 2 - 1$
$P(1) = 2 + 1 - 2 - 1 = 0$
Because $P(1) = 0$, $(y - 1)$ is definitively a valid factor. Similar checks for $y = -1$ and $y = -1/2$ will also yield $0$, confirming that all three linear binomials are the correct and complete factors of the given cubic polynomial.
Final Solution: The complete factorization of the polynomial $2y^3 + y^2 - 2y - 1$ is $(2y + 1)(y - 1)(y + 1)$.
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(ii): Factorise : (ii) $x^3 – 3x^2 – 9x – 5$
- Q5(iii): Factorise : (iii) $x^3 + 13x^2 + 32x + 20$
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
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