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Q4(iii):
Factorise :
(iii) $6x^2 + 5x – 6$
Solution :
Initial Setup & Given Polynomial
We are tasked with factorising the quadratic polynomial:
$P(x) = 6x^2 + 5x - 6$
This expression is in the standard quadratic form $ax^2 + bx + c$. To factorise it, we will employ the Splitting the Middle Term technique, also known as the AC Method [Per the fundamental principles of polynomial factorisation over integers].
Step 1: Coefficient Analysis & The AC Method
First, we identify the coefficients of the quadratic polynomial:
- Leading coefficient ($a$) = $6$
- Middle coefficient ($b$) = $5$
- Constant term ($c$) = $-6$
According to the AC Method, we must find the product of the leading coefficient and the constant term ($a \times c$):
$a \times c = 6 \times (-6) = -36$
Step 2: Splitting the Middle Term
We must now find two integers, let's call them $p$ and $q$, that satisfy two conditions simultaneously:
- Their product must equal $ac$: $p \times q = -36$
- Their sum must equal $b$: $p + q = 5$
Since the product ($-36$) is negative, the two numbers must have opposite signs. Since their sum ($5$) is positive, the number with the larger absolute value must be positive. Let us systematically evaluate the factor pairs of $36$:
| Factor Pair ($p, q$) | Product ($p \times q$) | Sum ($p + q$) | Condition Met? |
|---|---|---|---|
| $-1, 36$ | $-36$ | $35$ | No |
| $-2, 18$ | $-36$ | $16$ | No |
| $-3, 12$ | $-36$ | $9$ | No |
| $-4, 9$ | $-36$ | $5$ | Yes |
The correct integers are $9$ and $-4$. We will use these to split the middle term ($5x$).
Step 3: Algebraic Substitution and Grouping
Substitute $5x$ with $9x - 4x$ in the original polynomial:
$6x^2 + 9x - 4x - 6$
Next, we group the terms into pairs to factor out the Greatest Common Divisor (GCD) from each pair [Applying the Distributive Property $ab + ac = a(b+c)$]:
$= (6x^2 + 9x) - (4x + 6)$
Extract the GCD from the first group ($6x^2 + 9x$). The GCD of $6$ and $9$ is $3$, and the GCD of $x^2$ and $x$ is $x$. Thus, we factor out $3x$:
$= 3x(2x + 3) - (4x + 6)$
Extract the GCD from the second group ($4x + 6$). The GCD of $4$ and $6$ is $2$. To ensure the binomial inside the parentheses matches the first group, we factor out $-2$:
$= 3x(2x + 3) - 2(2x + 3)$
Step 4: Factoring out the Common Binomial
Notice that the binomial $(2x + 3)$ is now a common factor in both terms. We factor it out [By the reverse distributive property]:
$= (2x + 3)(3x - 2)$
Geometric Verification: Area Model
To rigorously verify our algebraic manipulation, we can map the grouped terms to a geometric area model. The total area of the rectangle represents the polynomial $6x^2 + 5x - 6$, while the side lengths represent its factors $(2x + 3)$ and $(3x - 2)$.
Final Solution: The factorised form of the polynomial $6x^2 + 5x - 6$ is $(2x + 3)(3x - 2)$.
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(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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