Let p(x) = x^4 + ax^3 + bx^2 + cx + d, where a, b, c, d are real numbers. It is known that...

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nanase

Junior Member
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Guys I am going in a loop for this one, may you give me a hint or suggestion on this :
polynomial.jpg
I tried creating equations from the information given, and hoping to solve them simultaneously as follows :
polynomial ans1.jpg
but I am stuck in only producing 1 equation which is 6a + b = -1025.
.
.
then I tried approaching from the last sentence information as follows :
polynomial ans2.jpeg
hoping I can see or obtain something, but I could not find/form anything useful.
Can you help me out please, thank you so much.
(I can't use calculator/technology for this question)
 
Since you have only three facts to solve for four unknowns, you can't determine all of them.

I would try to express all the parameters in terms of one (say, a), and then write an expression for what you are to find in terms of a, hoping that a will cancel out, and you'll get a fixed number. There may be a shortcut, but I don't see it yet. This should work, if you are careful enough.

I got 6a + b equal to a different number; I don't know which of us is wrong, but you'll want to check the details.
 
Just curious: in which class did you get this assignment? I.e., which topic are you studying there?
 
nanaseailie said:
Guys I am going in a loop for this one, may you give me a hint or suggestion on this :
View attachment 36704
I tried creating equations from the information given, and hoping to solve them simultaneously as follows :
View attachment 36705
but I am stuck in only producing 1 equation which is 6a + b = -1025.
.
.
then I tried approaching from the last sentence information as follows :
View attachment 36706
hoping I can see or obtain something, but I could not find/form anything useful.
Can you help me out please, thank you so much.
(I can't use calculator/technology for this question)
Click to expand...
You can get simpler equations if you look at [imath]q(x) = p(x+2)[/imath].
 
You've been asked before about your context; I have been assuming it is preparation for some sort of Olympiad, and (I now see) specifically the International Junior Math Olympiad (IJMO), judging by the background of a couple of your recent questions:

1699369351091.png​

In fact, I find almost your current problem (but not quite) here:


1699369605482.png​

Could you please tell us more about where the problems come from, and what background you have? And do you have any specific instruction in Olympiad solving methods?

It's hard to help someone without any idea what they have learned, and how far their questions are from their level of knowledge.
 
@blamocur & @DrPeterson, yes sir it is questions from various competitions I think such as SASMO, SIMOC, IJMO etc
I am a student in secondary school. I guess my level of Mathematics is a bit above average in my class, but I really like doing maths.
I am doing this not for the sake of submitting homework or anything urgent, I am just setting myself a goal to try doing some challenging & non text-book questions on regular basis when I am free. And I like to teach my friends, relatives and my lower grades friends in maths

The only specific instructions we have is no calculator/technology allowed.
 
Last edited:
blamocur said:
You can get simpler equations if you look at [imath]q(x) = p(x+2)[/imath].
Click to expand...
You mean I should make p(9) as p(7+2) or p(-5) as p(-7+2)?
Can you give a little more hint or initial steps pleaseee ?
 
nanaseailie said:
You mean I should make p(9) as p(7+2) or p(-5) as p(-7+2)?
Can you give a little more hint or initial steps pleaseee ?
Click to expand...
Re-state the original problem in terms of [imath]q(x)[/imath]. For example, for which [imath]x[/imath]'s do you know values of [imath]q[/imath]?
 
polynomial ans3.jpeg
I am continuously trying to reduce the equations with known information, but I still end up with a.
Please tell me is it miscalculation or my step is wrong?
 
I rechecked and I realised I made some mistakes in the calculation and able to rid off all the variables and solve for x
polynomial ans4.jpeg
but the answer is still wrong, now I am clueless why it is not correct
 
nanaseailie said:
you mean p(0) = d ?
Click to expand...
No. I'll rephrase my question : if [imath]q(x) = p(x+2)[/imath] for which values of [imath]x[/imath] do you know exact numeric values of [imath]q(x)[/imath] ? For example, if someone told me that [imath]p(100) = 1000[/imath], then I'd know that [imath]q(98) = 1000[/imath] because [imath]q(98) = p(2 + 98) = 1000[/imath].
 
blamocur said:
No. I'll rephrase my question : if [imath]q(x) = p(x+2)[/imath] for which values of [imath]x[/imath] do you know exact numeric values of [imath]q(x)[/imath] ? For example, if someone told me that [imath]p(100) = 1000[/imath], then I'd know that [imath]q(98) = 1000[/imath] because [imath]q(98) = p(2 + 98) = 1000[/imath].
Click to expand...
Can I get more hint on this please? I am confused because you are talking as if there are 2 different functions, but we only have one function which is p(x) right?

or can you give one example so I can see what I miss and reflect.
 
nanaseailie said:
I am confused because you are talking as if there are 2 different functions, but we only have one function which is p(x) right?
Click to expand...
Right. The problem mentions one function [imath]p[/imath], but there is nothing wrong (and occasionally quite useful) in introducing other functions during the solution.
Can you figure out the value of [imath]q(0)[/imath] ?
 
Just curious: why are you solving problems from a math olympiad? What are you trying to learn this way?
 
blamocur said:
Just curious: why are you solving problems from a math olympiad? What are you trying to learn this way?
Click to expand...
because I am good with textbooks / direct problems at my level, but when it comes to olympiad questions I sometimes feel blank. Since I love maths I would like to get better also in problem solving.

Right. The problem mentions one function pp, but there is nothing wrong (and occasionally quite useful) in introducing other functions during the solution.
Can you figure out the value of q(0)?


where can I sub the 0? what is q(x) ?
 
@blamocur & @Dr.Peterson I just realised something, the question I posted and what drPeterson posted is a little different. I tried working on DrPeterson question, I set up p(9) + p(-5) - 4 first which gives me 7182 + 604a + 106b + 4c + 2d.
Then I substituted 2xp(2) reducing the equation to 10514 + 588a + 98b.
from my earlier working of 6a +b = -1025 , I can further change the equation to -89936 only. Then divided this with -16 which gives me 5621.
Is this the correct answer?
The problem is my friend told me that 5621 is the answer for the question I posted, but he rigged the question or something, therefore I can't reach that solution.

Holy moly been busting my brain out and figuring out why things don't work out.
 
nanaseailie said:
I just realised something, the question I posted and what drPeterson posted is a little different.
Click to expand...
Yes, I said that:
Dr.Peterson said:
In fact, I find almost your current problem (but not quite) here:
Click to expand...
I wasn't saying you should solve that one, only that it suggested yours came from a related source. Of course, you can work on either one.

nanaseailie said:
The problem is my friend told me that 5621 is the answer for the question I posted, but he rigged the question or something, therefore I can't reach that solution.
Click to expand...
What do you mean by "rigged"?
 
polynomial-jpg.36704

It may help if we tell you the thinking behind the hint @blamocur gave.

You are given three points that lie on the graph of p: (1,841), (2,1682), and (3, 523). Plugging those into the equation and solving for a, b, c, and d takes a lot of work (and big numbers with many places to make mistakes). But with experience, we know that it would be easier to solve a problem like this if the values of x were close to zero, and -1, 0, 1 would be really helpful. That suggests shifting the graph left by 2, so the points become (-1,841), (0,1682), and (1, 523). These would be points on the graph of a new function q(x) = p(x+2).

Have you learned about transforming graphs? This is a "shift" or "translation" by 2 units to the left.

Now try doing the same sort of thing you were doing, with these new points, to find an equation of q(x). Then, for the final answer, you'll need to find p(9) and p(-5). What values of x do those correspond to on the graph of q? You'll find something interesting, which makes the problem even easier.

You've told us some things about yourself, but it may help more if you tell us what courses you have taken, and particularly what topics in algebra you have learned, so that we can know what sort of hints you will understand.

nanaseailie said:
I set up p(9) + p(-5) - 4 first which gives me 7182 + 604a + 106b + 4c + 2d.
Then I substituted 2xp(2) reducing the equation to 10514 + 588a + 98b.
from my earlier working of 6a +b = -1025 , I can further change the equation to -89936 only. Then divided this with -16 which gives me 5621.
Is this the correct answer?
The problem is my friend told me that 5621 is the answer for the question I posted, but he rigged the question or something, therefore I can't reach that solution.
Click to expand...
I don't have my work with me, and can't follow your description without seeing details. But you may be doing fine.
 
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