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Can you please help me with this maths question 1. The cubic polynomial P(x)= x^3 +rs^2 +sx+t , where r,s and t are real numbers has three real zeroes,1,a, and -a (a) find the value of s + t

Mathematics
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is the polynomial written correctly?
woops my bad sorry P(x) = x^3 +rx^2 +sx +t
Colorful pi can you help me?

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Other answers:

let me try to help him and then i will look at your question
him ? are yu referring to me ?
is this algebra?
noo , its a polynomial question
so what class is it then?
class ? what do yu mean ? Its from the topic polynomials
\[P(x)=(x-1)(x-a)(x+a)=x^3+rs^2+sx+t\]
is that the answer ?
so I guess you are doing it for fun? \[P(x)=(x-1)(x^2-a^2)=x^3+rs^2+sx+t\] no i just thought u would like to see the steps
no , im not doing it for fun , its my maths assignment LOL
\[P(x)=x(x^2-a^2)-1(x^2-a^2)=x^3+rsx^2+sx+t\]
so the class is called polynomials lol?
what does yu mean by class ?
\[P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t\] people go to class to learn and get assignments lol
no the class is called maths LOL
\[P(x)=x^3-x^2-a^2x+a^2=x^3+rsx^2+sx+t\]
2 unit maths to be specific
ok we are almost done with this problem can you guess what to do next?
we want both sides to be the same
collecting like terms ?
ok the the coefficient of x^2 on left hand side is -1 the coefficient of x^2 on the other sides is rs so what can you say about rs and -1?
they must be equal
so we need that for the other terms as well
we have -1=rs -a^2=s a^2=t
I seriously think I'm being trolled here.. but just to ease my tired brain.. \[\sqrt{25} \ne -5\]
you are right polpak
wait , dnt yu factorise it further to (x^2-a^2)(x-1)
that =5
factor?
Someone was trying to tell me that it equals both 5 and -5
ok the objective is to find s+t
But I can only really prove it by going to the definition. Which they don't accept.
ohh yeah , my bad
Sorry for bothering you mimi, but myin rarely watches group chat (with good reason)
what do you think s+t is based on what i gave you above? hint we don't need that first equation rs=-1
yu talking to me or polpak ? LOL
you
lols , tbh im lost , i dont get it xDD
do you understand when I wrote P as P(x)=(x-1)(x-a)(x+a)?
yeah , i dont understand the part -1 = rs -a^2 = s a^2 = t where did yu get it from ?
both sides had to be equal we were looking at \[P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t \]
look at the coefficients of each term on both sides we want each term to be the same as it is on the other side
so if we look at the x^3's they are good they both have coefficient 1 what is the coefficient of x^2 on left hand side ? what is the coefficient of x^2 on right hand side?
ohh i get it rs = -1
yes now what is the coefficient of x on the left hand side? what is the coefficient of x on the right hand side?
x on the left hand side: 1 i think x on the right hand side 1 i think
no what is the number in front of x on boths?
both sides*
there's no other number
-a^2? s?
lols , its a letter not a number
they are numbers
they are ? arent they are called pronumerals ?
are they*
never heard of this term
ohh ok
when you are doing math and you don't see a word and you see a letter it usually represents a number unless it is the word a or i
becauce a and i can be words
ohh ok , i never knew that they are numbers xD
but usually the book will make it clear to you what they are
yep
ok , so x= a^3
a^3 is nowhere to be found we aren;t solving equation for x
ok , so x= a^3
ohh ok
ok so look again at the polynomial what is the coefficent of x on both sides?
ok , so x= a^3
ok , so x= a^3
x= s
ok no \[P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t \] ^ ^ | | -a^2 = s
ok , so x= a^3
ohh lols , sorry im so slow of understanding it
ok do you see the terms that don't an x in em? there terms are called constants and they must be equal of that we have the same polynomial on both sides
god my english is getting bad
ok , so x= a^3
lols , probably getting annoyed of me not getting it xD
no its just late here i been up for along time today
a long*
Ditto!
so what are the constants on both sides?
the terms that don't have a x in them
ok , so x= a^3
t
ok , so x= a^3
t and what is it on the other side?
remember you are looking for non-x terms
ok , so x= a^3
a
a or a^2?
a
\[P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t \] ^ ^ | | a^2 = t
a^2 my bad
so what is s again? and what is t again?
s = 1 t = a^2
s is not 1
a^3
not a^3 scroll up we have it above
-a^2
lolss , im so~ slow
YES! :)
lols , finally i get it
ok now s+t is -a^2+a^2 right?
yeah
but what is -a^2+a^2?
so substitute it in s and t
idk
-5+5 is
-6+6 is -24+24 is -999+999 is
where did 5 come from ?
i'm just trying to help you think what -a^2+a^2 is
0 ?
yes! :)
and now you are done
really ? omg ! yay i got it right xD
thank you very mcuh for helping (:
np

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