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Q4(i):
Factorise :
(i) $12x^2 – 7x + 1$
Solution :
Initial Setup & Polynomial Identification
We are tasked with factorising the quadratic polynomial $P(x) = 12x^2 - 7x + 1$. [Per the Fundamental Theorem of Algebra and the properties of quadratic expressions, a polynomial of degree 2 can generally be factored into the product of two linear binomials]. We will utilize the method of splitting the middle term (also known as the AC method).
Step 1: Determine the Coefficients for the AC Method
A standard quadratic polynomial is expressed in the form $ax^2 + bx + c$. By comparing our given polynomial to the standard form, we identify the coefficients:
- $a = 12$ (Coefficient of $x^2$)
- $b = -7$ (Coefficient of $x$)
- $c = 1$ (Constant term)
Step 2: Find the Splitting Factors
To split the middle term, we must find two integers, let us define them as $p$ and $q$, that satisfy two specific conditions simultaneously:
- Product Condition: $p \times q = a \times c = 12 \times 1 = 12$
- Sum Condition: $p + q = b = -7$
[This algebraic condition ensures that the middle term is partitioned in a way that maintains the polynomial's equivalence while allowing for factorisation by grouping].
We list the factor pairs of $12$: $(1, 12)$, $(2, 6)$, and $(3, 4)$. Because the product ($+12$) is positive and the sum ($-7$) is negative, both factors $p$ and $q$ must be negative integers. Let us test the negative pairs:
- $(-1) + (-12) = -13$ (Incorrect)
- $(-2) + (-6) = -8$ (Incorrect)
- $(-3) + (-4) = -7$ (Correct)
Thus, the required splitting factors are $p = -4$ and $q = -3$.
Step 3: Rewrite the Polynomial by Splitting the Middle Term
We substitute the middle term $-7x$ with the equivalent expression $-4x - 3x$:
$P(x) = 12x^2 - 4x - 3x + 1$
Step 4: Factorise by Grouping
We now group the four terms into two distinct pairs to extract the Greatest Common Factor (GCF) from each pair:
$(12x^2 - 4x) - (3x - 1)$
Analyzing the first group: $(12x^2 - 4x)$
The highest common numerical factor of $12$ and $4$ is $4$. The highest common variable factor of $x^2$ and $x$ is $x$. Therefore, the GCF is $4x$.
Factoring out $4x$ yields: $4x(3x - 1)$
Analyzing the second group: $-(3x - 1)$
To ensure the binomial inside the parentheses matches the first group, we factor out $-1$.
Factoring out $-1$ yields: $-1(3x - 1)$
[By the Distributive Property of Multiplication over Addition, $ab + ac = a(b+c)$. We apply this in reverse to factor out the GCF].
Step 5: Extract the Common Binomial Factor
Substitute the factored groups back into the main expression:
$4x(3x - 1) - 1(3x - 1)$
Observe that the binomial $(3x - 1)$ is now a common factor to both major terms. We factor out $(3x - 1)$ from the entire expression:
$(3x - 1)(4x - 1)$
Final Solution: The factorised form of the polynomial $12x^2 - 7x + 1$ is $(4x - 1)(3x - 1)$.
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(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) $x^3 + 13x^2 + 32x + 20$
- 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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