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Q15(ii):
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (ii) Area : $35y^2 + 13y –12$

Solution :

Initial Setup & Theoretical Foundation

The geometric area of a rectangle is defined by the product of its two adjacent spatial dimensions, length and breadth. Mathematically, this is expressed as:

[$Area = Length \times Breadth$]

We are given the area of a rectangle as a quadratic polynomial in terms of the variable $y$:

$Area = 35y^2 + 13y - 12$

To find the possible expressions for the length and breadth, we must factorize this quadratic polynomial into the product of two linear binomials. Each resulting binomial will represent one of the dimensions of the rectangle.

Step 1: Coefficient Analysis and the Product-Sum Method

We will factorize the quadratic polynomial $ay^2 + by + c$ using the method of splitting the middle term. First, we identify the coefficients:

  • Leading coefficient ($a$) = $35$
  • Middle coefficient ($b$) = $13$
  • Constant term ($c$) = $-12$

According to the product-sum factorization theorem, we must find two real numbers, let's call them $p$ and $q$, such that:

1. Their product equals $a \times c$:
$p \times q = 35 \times (-12) = -420$

2. Their sum equals $b$:
$p + q = 13$

Step 2: Prime Factorization to Determine Split Values

To systematically find the values of $p$ and $q$, we analyze the prime factorization of the absolute value of the product ($420$):

$420 = 2 \times 210 = 2^2 \times 105 = 2^2 \times 3 \times 35 = 2^2 \times 3 \times 5 \times 7$

We need to group these prime factors into two numbers whose difference is $13$ (since the product is negative, one number must be positive and the other negative). Let us test combinations:

  • Combination 1: $(2^2 \times 5) = 20$ and $(3 \times 7) = 21$. Difference is $1$. (Incorrect)
  • Combination 2: $(2^2 \times 7) = 28$ and $(3 \times 5) = 15$. Difference is $13$. (Correct)

Since the sum must be positive ($+13$), the larger number must be positive. Therefore, our two numbers are $28$ and $-15$.

[$28 \times (-15) = -420$ and $28 + (-15) = 13$]

Step 3: Splitting the Middle Term and Grouping

We substitute the middle term $13y$ with $28y - 15y$ in the original polynomial:

$35y^2 + 28y - 15y - 12$

Next, we apply factorization by grouping. We group the first two terms and the last two terms:

$(35y^2 + 28y) - (15y + 12)$

Extract the greatest common divisor (GCD) from each group:

For $(35y^2 + 28y)$, the GCD is $7y$:
$7y(5y + 4)$

For $-(15y + 12)$, the GCD is $-3$:
$-3(5y + 4)$

Now, substitute these back into the expression:

$7y(5y + 4) - 3(5y + 4)$

Notice that the binomial $(5y + 4)$ is a common factor. Factoring it out yields the final product of two linear expressions:

$(7y - 3)(5y + 4)$

Step 4: Geometric Representation and Dimensional Assignment

Because multiplication is commutative [by the Commutative Property of Multiplication, $A \times B = B \times A$], either of these binomial factors can represent the length, and the other will represent the breadth.

Length: (7y - 3) Breadth: (5y + 4) Area = 35y² + 13y - 12

Final Solution: The possible expressions for the dimensions of the rectangle are Length = $(7y - 3)$ and Breadth = $(5y + 4)$, or vice versa.


More Questions from Class 9 Mathematics Polynomials EXERCISE 2.4


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