Q. 2 I

# Let’s resolve int

Answer :

(a3)2 + 3a3b3-40(b3)2

To resolve it into factors we first resolve 40

40=2×2×2×5

Given, x2 + (m + n)x + mn=(x + m)(x + n) … (2)

Comparing 1 and 2

m + n =3

mn =-40

(1) can be written as

(a3)2 + 8a3b3-5a3b3-40(b3)2

= (a3 + 8b3)( a3-5b3)

The resolved factors are (a3 + 8b3)( a3-5b3)

Apply the formula a3 + b3 = (a+b)(a2 + b2 – ab) in (a3 + 8b3)

Now (a3 + 8b3) = (a3 + (2b)3)

= (a2 + 4b2 – 2ab)( a3-5b3)

The resolved factors are (a2 + 4b2 – 2ab)( a3-5b3).

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses RELATED QUESTIONS :

I/we factorise thWest Bengal - Mathematics

I/we factorise thWest Bengal - Mathematics

I/we factorise thWest Bengal - Mathematics

I/we factorise thWest Bengal - Mathematics

Let’s resolve intWest Bengal - Mathematics

Let’s resolve theWest Bengal - Mathematics

Let’s resolve theWest Bengal - Mathematics

Let’s resolve intWest Bengal - Mathematics

Let’s resolve intWest Bengal - Mathematics

Let’s resolve intWest Bengal - Mathematics