# 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 $$=5 y+4$$

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