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Factorise: 16x5 - 144x3

 explanation..... see, here it's 16x5 - 144x3, so in these terms, 16 and x3 are common.

See this, you will understand, (2*2*2*2*x*x*x*x*x) - (2*2*2*2*3*3*x*x*x)

So the common among the two terms are 2*2*2*2*x*x*x = 16x3

Now, when you take16x3 as common, it will become like this,

16x3 (x2 - 9)

See, above this is given in the bracket (x2-9), for a while, forget about 16x3

You can also write it as (x2-32), right?

So, this matches with the identity, a2 - b2 = (a+b) (a-b)

Thus, (x2 - 32) which will be equal to (x+3) (x-3)

So, thus the factorised number will be 16x3 (x+3) (x-3)

Hope, you understood.... 

Well, is your book is of RS Agarwal ? 

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View Full Answer

16x5- 144x3 = x3 {(16x2 - 144)} = x3 {(4x)2 - (12)2}

Using the identity, a2-b2 = (a+b) (a-b) :

x3{(4x+12) (4x-12)}

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16 and x3 is common b/w these two terms.

so,

16x3(1x2-9)

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 please tell me why there is x3 {(16x2 - 144)} = x3 {(4x)2 - (12)2}

Using the identity, a2-b2 = (a+b) (a-b) :

x3{(4x+12) (4x-12)} after the equal to sign...???

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 Sorry for the mistake @Richa

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Sorry, I did a mistake 

This would be the correct answer, as Deepak has mentioned, but his answer also has a mistake.

16x5 - 144x3 

As 16 and x3 are common in both the terms, take it as common.

16x3 (x2 - 9) = 16x3 (x2 - 32) 

Using identity, a2 - b2 = (a+b) (a-b)

16x3 {(x+3) (x-3)}

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 i m telling u na....please explain the process fro where to get what??? i m not able 2 understand..dats y i didnt even understand that there was a mistake....

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 I am sorry if I was unable to make you understand. ;-(

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 ohh...it was so simple......thanku nov1699.. i have understood the solution of my problem noe....again thanku very much....and my book is of NCERT...why????

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 No thanks.. it's my duty to help... :-) 

No, I asked this because our book is of RS Agarwal and same question is there, along with that a4 + 2a2b2 + b4...... this question

So, bye Richa, now i have to go..... 

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