Question

1. Find the length and breadth of a rectangle whose area is (4a^4 + 28a^2 +45) square unit.​

Answer 1

Answer:

• Area = (4a⁴ + 28a² + 45) unit²
• Find Length and Breadth of the rectangle

We know that Area of rectangle is (Length × Breadth), so need to factorize the area in that form only.

• According to the Question :

⇒ Area = (4a⁴ + 28a² + 45)

(18 + 10) = 28

• (18 × 10) = (4 × 45) = 180

⇒ Length × Breadth = 4a⁴ + (18 + 10)a² + 45

⇒ L × B = 4a⁴ + 18a² + 10a² + 45

⇒ L × B = 2a²(2a² + 9) + 5(2a² + 9)

⇒ L × B = (2a² + 5)(2a² + 9)

• By comparing both

⇒ Length = (2a² + 5) and,

⇒ Breadth = (2a² + 9)

∴ The length and breadth of rectangle are (2a² + 5) and (2a² + 9) respectively.

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How to do factorisation?

• To factorise a quadratic equation (x² + bx + c) we have to find numbers p and q such that p + q = b and pq = c.
• After finding p and q, we split the middle term in the quadratic as px + qx and get desired factors by grouping the terms.

Here is an Example :

Factorize x² + 6x + 8

To factorize x² + 6x + 8, we find two numbers p and q such that p + q = 6 and pq = 8.

Clearly, 2 + 4 = 6 and 2 × 4 = 8.

We know split the middle term 6x in the given quadratic as (2x + 4x), so that

⇢ x² + 6x + 8

⇢ x² + 2x + 4x + 8

⇢ (x² + 2x) + (4x + 8)

⇢ x(x + 2) + 4(x+ 2)

⇢ (x + 2)(x + 4)

Answer 2

the answer is

$(2a {}^{2} + 5)(2a {}^{2} + 9)$

it's easy