My function doubts!

  • Iff(x+3)=x2+x6,thenfindf(x).If\quad f\left( x+3 \right) \quad =\quad { x }^{ 2 }+x-6,\quad then\quad find\quad f(x).\quad

  • Iff(x+3)=x27x+2,findtheremainderwhenf(x)isdividedbyx+1.If\quad f(x+3)\quad =\quad { x }^{ 2 }-7x+2,\quad find\quad the\quad remainder\quad when\quad f(x)\quad is\quad divided\quad by\quad x+1.

Please help me with the problems, Thanks!

#Algebra

Note by Swapnil Das
6 years ago

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Comments

They are very easy.

1.Let g(x) = f(x), Not let y = x+3, we getg(y)=(y3)2+(y3)6f(x)=x25x\text{Let g(x) = f(x), Not let y = x+3, we get} \\ g(y) = (y-3)^2 + (y-3) - 6 \\ f(x) = x^2 - 5x

2.We know that the remainder will be f(-1). We have to find f(-1) We know that f(x+3)=x27x+2Now if we let x = -4 we’ll get f(-1) hence our answer.Hence f(-1) will bef(1)=(4)27(4)+2=46\text{We know that the remainder will be f(-1). We have to find f(-1) } \\ \text{We know that } f(x+3) = x^2 -7x +2 \\ \text{Now if we let x = -4 we'll get f(-1) hence our answer.} \\ \text{Hence f(-1) will be} f(-1) = (-4)^2 - 7(4) + 2 = 46

Rajdeep Dhingra - 6 years ago

@Swapnil Das @Sravanth Chebrolu I think you must revise the simple remainder- theorem for the 2nd question.

Nihar Mahajan - 6 years ago

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Yup, I have. I just got a last doubt in the chapter, so I asked for help. Well I was now solving problems of Newton's Sums. @Nihar Mahajan

Swapnil Das - 6 years ago

let, x+3 =y......

make a new function.....

and solve....

parth tandon - 6 years ago

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i think i am little slow in typing.......:p

parth tandon - 6 years ago

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Why do you think so?

Sravanth C. - 6 years ago

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@Sravanth C. when i started posting the solution there were no comments but as soon as i finished i found 6 comments hovering around....:)

parth tandon - 6 years ago

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@Parth Tandon Whoah! That happens to me too! But, don't consider that as your inability.

Sravanth C. - 6 years ago

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@Sravanth C. ya.........

i am working on it......

always used to work with a pen and paper so..................

parth tandon - 6 years ago

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@Parth Tandon Oh yes! The classic method of pen and paper . . . :P

Sravanth C. - 6 years ago

  • First question: the answer is x25xx^{2}-5x

  • Second question: the answer is 4646.

Am I right? If I am I'll post the solution. Please reply @Swapnil Das

Sravanth C. - 6 years ago

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Ur answer to 2nd one is wrong. For a level 5 like U it should Child's play. Anyways u can see my solution. Cheers!

Rajdeep Dhingra - 6 years ago

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Sorry, I got it wrong as I got meddled up with the signs, I posted it correctly on Swapnil's message board though, cheers! xD

Sravanth C. - 6 years ago

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@Sravanth C. Yeah!

Swapnil Das - 6 years ago

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@Swapnil Das Hey! I posted a question on logic just now, as you wanted me to post such questions I did it, here's the link!

Sravanth C. - 6 years ago

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@Sravanth C. See who have solved...

Swapnil Das - 6 years ago

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@Swapnil Das Omg! That was qiuck! Hope you liked it. . .

Sravanth C. - 6 years ago

I think you are right. Please post the solution.

Swapnil Das - 6 years ago

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But I think it's already posted! :P

Sravanth C. - 6 years ago

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@Sravanth C. Well Rajdeep also did the same, BTW thank you very much. Just a request, if you have time anytime, please post problems of this format, I would surely like them!

Swapnil Das - 6 years ago

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@Swapnil Das

We'll take that into consideration. :P

Surely! I'll keep posting suchquestions!

Sravanth C. - 6 years ago

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@Sravanth C. Go to Community-All topics-Problems

Swapnil Das - 6 years ago

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@Swapnil Das Okay. So that's nothing but explore?

Sravanth C. - 6 years ago

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@Sravanth C. yeah!

Swapnil Das - 6 years ago

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@Swapnil Das Thank you very much! BTW, how did you like the new look?

Sravanth C. - 6 years ago

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@Sravanth C. Nice!

Swapnil Das - 6 years ago

Can you please tell me how it became -5y instead of -5x?

Swapnil Das - 6 years ago

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Sorry, that was a typo there! It's edited now.

Sravanth C. - 6 years ago

BTW, I liked the new look of Brilliant! But I didn't find the explore icon anywhere. . .

Sravanth C. - 6 years ago

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@Sravanth C. May I help U?

Swapnil Das - 6 years ago

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@Swapnil Das anyone.....?

multi step decryption....?

parth tandon - 6 years ago

@Sravanth C. community.............

parth tandon - 6 years ago

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@Parth Tandon Not the community, the was an icon explore, I didn't find that...

Sravanth C. - 6 years ago

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@Sravanth C. u can check community ....................

i think explore renamed to community................

latest posts are there which before were given under explore................

parth tandon - 6 years ago

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@Parth Tandon multi step decryption...............?

parth tandon - 6 years ago

@Parth Tandon Yes. I saw that. Thank you very much!

Sravanth C. - 6 years ago

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@Sravanth C. welcome...............

do u know what is multi step decryption or i hv to search on net.....?

parth tandon - 6 years ago

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@Parth Tandon I did hear about that in a novel Digital Fortress, but I don't know much about it. . . :(

Sravanth C. - 6 years ago

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@Sravanth C. ok then........................

thnx for giving a thought to it..................

parth tandon - 6 years ago

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@Parth Tandon Welcome!

Sravanth C. - 6 years ago

Hey Sravanth, I don't know how I am getting some other remainder. Can you post the 2nd solution please?

Swapnil Das - 6 years ago

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Sure. Here it is.

As you know how to make expression, you'll get f(x):x213x10f(x):x^2-13x-10

Dividing it by x+1x+1, we get 44. . .

Sravanth C. - 6 years ago

can someone tell me what is multi_step decryption.....?

parth tandon - 6 years ago
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