Iff(x+3)=x2+x−6,thenfindf(x).If\quad f\left( x+3 \right) \quad =\quad { x }^{ 2 }+x-6,\quad then\quad find\quad f(x).\quadIff(x+3)=x2+x−6,thenfindf(x).
Iff(x+3)=x2−7x+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.Iff(x+3)=x2−7x+2,findtheremainderwhenf(x)isdividedbyx+1.
Please help me with the problems, Thanks!
Note by Swapnil Das 6 years ago
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They are very easy.
1.Let g(x) = f(x), Not let y = x+3, we getg(y)=(y−3)2+(y−3)−6f(x)=x2−5x\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 - 5xLet g(x) = f(x), Not let y = x+3, we getg(y)=(y−3)2+(y−3)−6f(x)=x2−5x
2.We know that the remainder will be f(-1). We have to find f(-1) We know that f(x+3)=x2−7x+2Now if we let x = -4 we’ll get f(-1) hence our answer.Hence f(-1) will bef(−1)=(−4)2−7(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 We know that the remainder will be f(-1). We have to find f(-1) We know that f(x+3)=x2−7x+2Now if we let x = -4 we’ll get f(-1) hence our answer.Hence f(-1) will bef(−1)=(−4)2−7(4)+2=46
@Swapnil Das @Sravanth Chebrolu I think you must revise the simple remainder- theorem for the 2nd question.
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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
let, x+3 =y......
make a new function.....
and solve....
i think i am little slow in typing.......:p
Why do you think so?
@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 – Whoah! That happens to me too! But, don't consider that as your inability.
@Sravanth C. – ya.........
i am working on it......
always used to work with a pen and paper so..................
@Parth Tandon – Oh yes! The classic method of pen and paper . . . :P
Second question: the answer is 464646.
Am I right? If I am I'll post the solution. Please reply @Swapnil Das
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!
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. – Yeah!
@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. – See who have solved...
@Swapnil Das – Omg! That was qiuck! Hope you liked it. . .
I think you are right. Please post the solution.
But I think it's already posted! :P
@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 –
We'll take that into consideration. :P
Surely! I'll keep posting suchquestions!
@Sravanth C. – Go to Community-All topics-Problems
@Swapnil Das – Okay. So that's nothing but explore?
@Sravanth C. – yeah!
@Swapnil Das – Thank you very much! BTW, how did you like the new look?
@Sravanth C. – Nice!
Can you please tell me how it became -5y instead of -5x?
Sorry, that was a typo there! It's edited now.
BTW, I liked the new look of Brilliant! But I didn't find the explore icon anywhere. . .
@Sravanth C. – May I help U?
@Swapnil Das – anyone.....?
multi step decryption....?
@Sravanth C. – community.............
@Parth Tandon – Not the community, the was an icon explore, I didn't find that...
@Sravanth C. – u can check community ....................
i think explore renamed to community................
latest posts are there which before were given under explore................
@Parth Tandon – multi step decryption...............?
@Parth Tandon – Yes. I saw that. Thank you very much!
@Sravanth C. – welcome...............
do u know what is multi step decryption or i hv to search on net.....?
@Parth Tandon – I did hear about that in a novel Digital Fortress, but I don't know much about it. . . :(
@Sravanth C. – ok then........................
thnx for giving a thought to it..................
@Parth Tandon – Welcome!
Hey Sravanth, I don't know how I am getting some other remainder. Can you post the 2nd solution please?
Sure. Here it is.
As you know how to make expression, you'll get f(x):x2−13x−10f(x):x^2-13x-10f(x):x2−13x−10
Dividing it by x+1x+1x+1, we get 444. . .
can someone tell me what is multi_step decryption.....?
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Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
They are very easy.
1.Let g(x) = f(x), Not let y = x+3, we getg(y)=(y−3)2+(y−3)−6f(x)=x2−5x
2.We know that the remainder will be f(-1). We have to find f(-1) We know that f(x+3)=x2−7x+2Now if we let x = -4 we’ll get f(-1) hence our answer.Hence f(-1) will bef(−1)=(−4)2−7(4)+2=46
@Swapnil Das @Sravanth Chebrolu I think you must revise the simple remainder- theorem for the 2nd question.
Log in to reply
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
let, x+3 =y......
make a new function.....
and solve....
Log in to reply
i think i am little slow in typing.......:p
Log in to reply
Why do you think so?
Log in to reply
Log in to reply
Log in to reply
i am working on it......
always used to work with a pen and paper so..................
Log in to reply
Second question: the answer is 46.
Am I right? If I am I'll post the solution. Please reply @Swapnil Das
Log in to reply
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!
Log in to reply
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
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link!
Hey! I posted a question on logic just now, as you wanted me to post such questions I did it, here's theLog in to reply
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I think you are right. Please post the solution.
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But I think it's already posted! :P
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Surely! I'll keep posting suchquestions!
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Can you please tell me how it became -5y instead of -5x?
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Sorry, that was a typo there! It's edited now.
BTW, I liked the new look of Brilliant! But I didn't find the explore icon anywhere. . .
Log in to reply
Log in to reply
multi step decryption....?
Log in to reply
Log in to reply
i think explore renamed to community................
latest posts are there which before were given under explore................
Log in to reply
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do u know what is multi step decryption or i hv to search on net.....?
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thnx for giving a thought to it..................
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Hey Sravanth, I don't know how I am getting some other remainder. Can you post the 2nd solution please?
Log in to reply
Sure. Here it is.
As you know how to make expression, you'll get f(x):x2−13x−10
Dividing it by x+1, we get 4. . .
can someone tell me what is multi_step decryption.....?