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jiteshmeghwal9
Group Title
Can anyone teach me about HCF of two polynomials ??
 2 years ago
 2 years ago
jiteshmeghwal9 Group Title
Can anyone teach me about HCF of two polynomials ??
 2 years ago
 2 years ago

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ash2326 Group TitleBest ResponseYou've already chosen the best response.2
Yeah. I can explain. Do you know how to find HCF of two no.s ?
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
yes !
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
Great :) Works the same way. Suppose I have two polynomials \[x^2+4x+4\ \text{and}\ {x^2+5x+6}\]
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
We need to find the highest common factor of these two. First step is to factor the polynomial (obviously only if they could be factored ). Can you factor these polynomials ?
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
okay !
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
x^2+2x+2x+4 x(x+2)+2(x+2) (x+2)^2 x^2+5x+6 x^2+3x+2x+6 x(x+3)+2(x+3) (x+2)(x+3)
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
Do you see any common factor in the polynomials ?
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
(x+2)
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
Yes, this is the only common factor and this is our HCF
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
If you have multiple common factors, just multiply all the common factors and that is your HCF
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
Do you understand this?
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
Ohh ! Thanx @ash2326 u were of a gr8 help :)
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
You're welcome :)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
yes. good explanation @ash2326 :)
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
Thanks @hartnn
 2 years ago
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