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

- Mimi_x3

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- myininaya

is the polynomial written correctly?

- Mimi_x3

woops my bad sorry
P(x) = x^3 +rx^2 +sx +t

- helpme1234

Colorful pi can you help me?

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## More answers

- myininaya

let me try to help him and then i will look at your question

- Mimi_x3

him ? are yu referring to me ?

- myininaya

is this algebra?

- Mimi_x3

noo , its a polynomial question

- myininaya

so what class is it then?

- Mimi_x3

class ? what do yu mean ? Its from the topic polynomials

- myininaya

\[P(x)=(x-1)(x-a)(x+a)=x^3+rs^2+sx+t\]

- Mimi_x3

is that the answer ?

- myininaya

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

- Mimi_x3

no , im not doing it for fun , its my maths assignment LOL

- myininaya

\[P(x)=x(x^2-a^2)-1(x^2-a^2)=x^3+rsx^2+sx+t\]

- myininaya

so the class is called polynomials lol?

- Mimi_x3

what does yu mean by class ?

- myininaya

\[P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t\]
people go to class to learn and get assignments lol

- Mimi_x3

no the class is called maths LOL

- myininaya

\[P(x)=x^3-x^2-a^2x+a^2=x^3+rsx^2+sx+t\]

- Mimi_x3

2 unit maths to be specific

- myininaya

ok we are almost done with this problem
can you guess what to do next?

- myininaya

we want both sides to be the same

- Mimi_x3

collecting like terms ?

- myininaya

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?

- myininaya

they must be equal

- myininaya

so we need that for the other terms as well

- myininaya

we have
-1=rs
-a^2=s
a^2=t

- anonymous

I seriously think I'm being trolled here.. but just to ease my tired brain..
\[\sqrt{25} \ne -5\]

- myininaya

you are right polpak

- Mimi_x3

wait , dnt yu factorise it further to (x^2-a^2)(x-1)

- myininaya

that =5

- myininaya

factor?

- anonymous

Someone was trying to tell me that it equals both 5 and -5

- myininaya

ok the objective is to find s+t

- anonymous

But I can only really prove it by going to the definition. Which they don't accept.

- Mimi_x3

ohh yeah , my bad

- anonymous

Sorry for bothering you mimi, but myin rarely watches group chat (with good reason)

- myininaya

what do you think s+t is based on what i gave you above?
hint we don't need that first equation rs=-1

- Mimi_x3

yu talking to me or polpak ? LOL

- myininaya

you

- Mimi_x3

lols , tbh im lost , i dont get it xDD

- myininaya

do you understand when I wrote P
as P(x)=(x-1)(x-a)(x+a)?

- Mimi_x3

yeah , i dont understand the part
-1 = rs
-a^2 = s
a^2 = t
where did yu get it from ?

- myininaya

both sides had to be equal we were looking at
\[P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t \]

- myininaya

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

- myininaya

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?

- Mimi_x3

ohh i get it
rs = -1

- myininaya

yes
now
what is the coefficient of x on the left hand side?
what is the coefficient of x on the right hand side?

- Mimi_x3

x on the left hand side: 1 i think
x on the right hand side 1 i think

- myininaya

no what is the number in front of x on boths?

- myininaya

both sides*

- Mimi_x3

there's no other number

- myininaya

-a^2?
s?

- Mimi_x3

lols , its a letter not a number

- myininaya

they are numbers

- Mimi_x3

they are ? arent they are called pronumerals ?

- Mimi_x3

are they*

- myininaya

never heard of this term

- Mimi_x3

ohh ok

- myininaya

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

- myininaya

becauce a and i can be words

- Mimi_x3

ohh ok , i never knew that they are numbers xD

- myininaya

but usually the book will make it clear to you what they are

- myininaya

yep

- Mimi_x3

ok , so x= a^3

- myininaya

a^3 is nowhere to be found
we aren;t solving equation for x

- Mimi_x3

ok , so x= a^3

- Mimi_x3

ohh ok

- myininaya

ok so look again at the polynomial
what is the coefficent of x on both sides?

- Mimi_x3

ok , so x= a^3

- Mimi_x3

ok , so x= a^3

- Mimi_x3

x= s

- myininaya

ok no
\[P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t \]
^ ^
| |
-a^2 = s

- Mimi_x3

ok , so x= a^3

- Mimi_x3

ohh lols , sorry im so slow of understanding it

- myininaya

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

- myininaya

god my english is getting bad

- Mimi_x3

ok , so x= a^3

- Mimi_x3

lols , probably getting annoyed of me not getting it xD

- myininaya

no its just late here
i been up for along time today

- myininaya

a long*

- anonymous

Ditto!

- myininaya

so what are the constants on both sides?

- myininaya

the terms that don't have a x in them

- Mimi_x3

ok , so x= a^3

- Mimi_x3

t

- Mimi_x3

ok , so x= a^3

- myininaya

t and what is it on the other side?

- myininaya

remember you are looking for non-x terms

- Mimi_x3

ok , so x= a^3

- Mimi_x3

a

- myininaya

a or a^2?

- Mimi_x3

a

- myininaya

\[P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t \]
^ ^
| |
a^2 = t

- Mimi_x3

a^2 my bad

- myininaya

so what is s again?
and what is t again?

- Mimi_x3

s = 1
t = a^2

- myininaya

s is not 1

- Mimi_x3

a^3

- myininaya

not a^3
scroll up
we have it above

- Mimi_x3

-a^2

- Mimi_x3

lolss , im so~ slow

- myininaya

YES! :)

- Mimi_x3

lols , finally i get it

- myininaya

ok now s+t is -a^2+a^2 right?

- Mimi_x3

yeah

- myininaya

but what is -a^2+a^2?

- Mimi_x3

so substitute it in s and t

- Mimi_x3

idk

- myininaya

-5+5 is

- myininaya

-6+6 is
-24+24 is
-999+999 is

- Mimi_x3

where did 5 come from ?

- myininaya

i'm just trying to help you think what -a^2+a^2 is

- Mimi_x3

0 ?

- myininaya

yes! :)

- myininaya

and now you are done

- Mimi_x3

really ? omg ! yay i got it right xD

- Mimi_x3

thank you very mcuh for helping (:

- myininaya

np

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