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by killer1387 » Fri Mar 23, 2012 11:15 pm
romitvsingh wrote:the LCM of polynomials (36x^2-49) and (6x^2-25x+21)is

a (x-3)

b (x-3)^2 (6x-7)

c (x-3) (6x-7) (6x+7)

d none of these
(36x^2-49) = (6x-7)*(6x+7), and
(6x^2-25x+21)= (6x-7)*(x-3)

LCM=(X-3)*(6X+7)*(6X-7)

HENCE C
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by [email protected] » Sun Apr 01, 2012 3:30 am
Well bro killer1387 you have committed a minor mistake...


The answer should be D and not C. the reason is that see the factors once...\

the factors of 6x^2 - 9x - 14x + 21

= 3x (2x - 3) - 7 (2x - 3)

= (3x - 7) (2x - 3)

hence the factors are (3x - 7) (2x - 3) (6x - 7) (6x + 7)


none of the options A , B or C are the above hence D.

I am sure of this... please see this and let me know


By the way what is the OA romitvsingh...
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by Pharo » Sun Apr 01, 2012 3:39 am
[email protected] wrote:Well bro killer1387 you have committed a minor mistake...


The answer should be D and not C. the reason is that see the factors once...\

the factors of 6x^2 - 9x - 14x + 21

= 3x (2x - 3) - 7 (2x - 3)

= (3x - 7) (2x - 3)

hence the factors are (3x - 7) (2x - 3) (6x - 7) (6x + 7)


none of the options A , B or C are the above hence D.

I am sure of this... please see this and let me know


By the way what is the OA romitvsingh...
Nope, you made a mistake: 9 + 14 = 23; not 25 :)

The answer is C.
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by Anurag@Gurome » Sun Apr 01, 2012 8:53 pm
romitvsingh wrote:the LCM of polynomials (36x^2-49) and (6x^2-25x+21)is

a (x-3)

b (x-3)^2 (6x-7)

c (x-3) (6x-7) (6x+7)

d none of these
(36x^2 - 49) = (6x + 7)(6x - 7)
(6x^2 - 25x + 21) = 6x^2 - 18x - 7x + 21 = 6x(x - 3) - 7(x - 3) = (x - 3)(6x - 7)

Therefore, LCM of (36x^2 - 49) and (6x^2 - 25x + 21) = [spoiler](x - 3)(6x + 7)(6x - 7) [/spoiler]

The correct answer is C.
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by Anurag@Gurome » Sun Apr 01, 2012 8:53 pm
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Last edited by Anurag@Gurome on Sun Apr 01, 2012 8:54 pm, edited 1 time in total.
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by Anurag@Gurome » Sun Apr 01, 2012 8:54 pm
Delete.
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by [email protected] » Mon Apr 02, 2012 10:40 pm
yes thank you guyzzz a silly mistake from my side... I am sorry.. I got my mistake...
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