Ex.2.5 Q15 Polynomials Solution - NCERT Maths Class 9

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Question

Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

(i) Area: \(25 a^{2}-35 a+12\)                                         

(ii) Area: \(35 y^{2}+13 y-12\)

 

 Video Solution
Polynomials
Ex 2.5 | Question 15

Text Solution

  

Reasoning:

\[\text{Area of rectangle} =\text {Length } \times \text {Breadth}\]

What is known?

Area of rectangle

What is  unknown?

Length and breadth of the rectangle.

Steps:

\[\text{Area of rectangle} =\text {Length } \times \text {Breadth}\]

Hence, we shall factorise the given expression \(25 a^{2}-35 a+12\)

Now taking  \(25 a^{2}-35 a+12\)  find \(2\) numbers \(p, q\) such that:

(i) \(p+q=\)  co-efficient of \(a\)  

(ii) \(p q=\)  co-efficient of \(a^{2}\) and the constant term.

\(p+q=-35\)  (co-efficient of a ) 

\(p q=25 \times 12=300\)  (co-efficient of \(a^{2}\) and the constant term.) 

By trial and error method, we get \(p=-20, q=-15\)

Now splitting the middle term of the given polynomial,

\[\begin{align}25 a^{2}-35 a+12 &=25 a^{2}-20 a-15 a+12 \\ &=25 a^{2}-15 a-20 a+12 \\ &=5 a(5 a-3)-4(5 a-3) \\ &=(5 a-4)(5 a-3) \end{align}\]

\(\begin{align}\therefore 25 a^{2}-35 a+12 &=(5 a-4)(5 a-3)\end{align}\)

Length\(=\)\(5 a-3\)  Breadth\(=5 a-4\)

Length\(=5 a-4\) Breadth\(=5 a-3\)

What is known?

Area of rectangle.

What is unknown?

 Length and breadth of the rectangle.

Steps:

\[\text{Area of rectangle} =\text {Length } \times \text {Breadth}\]

Hence, we shall factorise the given expression: \(35 y^{2}+13 y-12\)

Now taking \(35 y^{2}-13 y-12\) , find \(2\) numbers \(p, q\) such that:

i. \(p+q=\) co-efficient of \(y\)

ii. \(p q=\)co-efficient of \(y^{2}\)  and the constant term.

\begin{align}&{p+q}={-13 \text { (co-efficient of } y )} \\ &{p q}={35 \times-12=-420\left(\text { co-efficient of } y^{2} \text { and the constant term.) }\right.}\end{align}

By trial and error method, we get \( p = -28, q = -15.\)

Now splitting the middle term of the given polynomial,

\[\begin{align} 35 y^{2}+13 y-12 &=35 y^{2}+28 y-15 y-12 \\ &=7 y(5 y+4)-3(5 y+4) \\ &=(5 y+4)(7 y-3)\end{align}\]

\(\begin{align} \therefore 35 y^{2}+13 y-12 &=(5 y+4)(7 y-3)\end{align}\)

 Length \(=5 y+4\)  Breadth \(=7 y-3\) 

Length\(=7 y-3\) Breadth \(=5 y+4\)