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Easy remainder theorem problem

x4 + x3 + x - 10 = (ax2 + bx + c) (x2 + 2x - 3) + Dx + e

I am supposed to solve for a,b,c,d and e.

i think the first step is to sub in -3 where ever there is an x to find d?

Thank you!!
Original post by JamesNeedHelp2
x4 + x3 + x - 10 = (ax2 + bx + c) (x2 + 2x - 3) + Dx + e

I am supposed to solve for a,b,c,d and e.

i think the first step is to sub in -3 where ever there is an x to find d?

Thank you!!

You could indeed find out −3d+e-3d+e that way, and if you substitute x=1x=1 you'll get another expression which is d+ed+e. Those you can solve simultaneously.

To get the a,b,c terms, it's easiest just to expand the brackets of the right-hand side and compare coefficients, in my opinion.
Reply 2
Avatar for davros
davros
16
Original post by JamesNeedHelp2
x4 + x3 + x - 10 = (ax2 + bx + c) (x2 + 2x - 3) + Dx + e

I am supposed to solve for a,b,c,d and e.

i think the first step is to sub in -3 where ever there is an x to find d?

Thank you!!


I presume the numbers after the x's are supposed to be indices?

How far have you got - you should be able to write down the value of a straight away just by looking!
Original post by Smaug123
You could indeed find out −3d+e-3d+e that way, and if you substitute x=1x=1 you'll get another expression which is d+ed+e. Those you can solve simultaneously.

To get the a,b,c terms, it's easiest just to expand the brackets of the right-hand side and compare coefficients, in my opinion.


Thank you sir.:smile:
Original post by davros
I presume the numbers after the x's are supposed to be indices?

How far have you got - you should be able to write down the value of a straight away just by looking!


Is it true that i am supposed to compare the coefficients of x to find a? if so, i think a= 1?
Original post by JamesNeedHelp2
Is it true that i am supposed to compare the coefficients of x to find a? if so, i think a= 1?


Quite correct.
Reply 6
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davros
16
Original post by JamesNeedHelp2
Is it true that i am supposed to compare the coefficients of x to find a? if so, i think a= 1?


You can compare coefficients to get simultaneous equations that will give you all the unknown constants; you can also do things like sub in various x values like 1 and -3 which I think will make one of your brackets vanish and let you find d and e quite simply :smile:
Original post by davros
You can compare coefficients to get simultaneous equations that will give you all the unknown constants; you can also do things like sub in various x values like 1 and -3 which I think will make one of your brackets vanish and let you find d and e quite simply :smile:



Thanks again!!:cool:
Original post by brianeverit
Quite correct.


Thank you!!:smile:
Original post by Smaug123
You could indeed find out −3d+e-3d+e that way, and if you substitute x=1x=1 you'll get another expression which is d+ed+e. Those you can solve simultaneously.

To get the a,b,c terms, it's easiest just to expand the brackets of the right-hand side and compare coefficients, in my opinion.


Am i over thinking it or is it really difficult to find -3d + e?

So i subbed in -3;

(a(-3)2 - 3b + c) ((-3)2 + 2(-3) - 3) + dx +e

9 - 3b + c -3d + e

Is this correct? if so, and how do i proceed from here to find -3d + e?

Thank you!!
Reply 10
Avatar for TeeEm
TeeEm
19
Original post by JamesNeedHelp2
Am i over thinking it or is it really difficult to find -3d + e?

So i subbed in -3;

(a(-3)2 - 3b + c) ((-3)2 + 2(-3) - 3) + dx +e

9 - 3b + c -3d + e

Is this correct? if so, and how do i proceed from here to find -3d + e?

Thank you!!


I try to avoid coming into a thread where others (perhaps too many) are trying to help you, but ....

... is there something wrong in doing a long division which takes 30 or so seconds and read off the answers for all the constants?
Original post by TeeEm
I try to avoid coming into a thread where others (perhaps too many) are trying to help you, but ....

... is there something wrong in doing a long division which takes 30 or so seconds and read off the answers for all the constants?


I didnt really know that was possible. There seems to be quite a few methods, and i am baffled by all of them. But i will try this method out.:smile:
Reply 12
Avatar for TeeEm
TeeEm
19
Original post by JamesNeedHelp2
I didnt really know that was possible. There seems to be quite a few methods, and i am baffled by all of them. But i will try this method out.:smile:


in my humble opinion what you are being told is correct but certainly a quick long division is the most effective method
Original post by TeeEm
in my humble opinion what you are being told is correct but certainly a quick long division is the most effective method


Thank you. In this problem, x4 + x3 + x - 10 = (ax2 + bx + c) (x2 + 2x - 3) + Dx + e, am i dividing x4 + x3 + x - 10 by ax2 + bx + c?

I am just trying to figure out, what i need to divide by what to find the constants a, b, c, d an e?

Thanks again!
Reply 14
Avatar for TeeEm
TeeEm
19
Original post by JamesNeedHelp2
Thank you. In this problem, x4 + x3 + x - 10 = (ax2 + bx + c) (x2 + 2x - 3) + Dx + e, am i dividing x4 + x3 + x - 10 by ax2 + bx + c?

I am just trying to figure out, what i need to divide by what to find the constants a, b, c, d an e?

Thanks again!


x4 + x3 + x - 10 by (x2 + 2x - 3)
Original post by JamesNeedHelp2
Am i over thinking it or is it really difficult to find -3d + e?

So i subbed in -3;

(a(-3)2 - 3b + c) ((-3)2 + 2(-3) - 3) + dx +e

9 - 3b + c -3d + e

Is this correct? if so, and how do i proceed from here to find -3d + e?

Thank you!!

(−3)2+2×(−3)−3(-3)^2 + 2 \times (-3) - 3 is 0. That entire bracket vanishes.
Original post by Smaug123
(−3)2+2×(−3)−3(-3)^2 + 2 \times (-3) - 3 is 0. That entire bracket vanishes.


ok. Which leaves me with 9 - 3b + c -3d + e:confused:
Original post by davros
You can compare coefficients to get simultaneous equations that will give you all the unknown constants; you can also do things like sub in various x values like 1 and -3 which I think will make one of your brackets vanish and let you find d and e quite simply :smile:



Original post by JamesNeedHelp2
Am i over thinking it or is it really difficult to find -3d + e?

So i subbed in -3;

(a(-3)2 - 3b + c) ((-3)2 + 2(-3) - 3) + dx +e

9 - 3b + c -3d + e

Is this correct? if so, and how do i proceed from here to find -3d + e?

Thank you!!



So baffled :biggrin:
Reply 18
Avatar for davros
davros
16
Original post by JamesNeedHelp2
ok. Which leaves me with 9 - 3b + c -3d + e:confused:


No it doesn't!

If one of the brackets is 0 then you just get 0 for the 1st term on the RHS leaving you with -3d + e.

You can plug in x = 1 to get a similar equation, or plug in lots of other small values of x to get a set of simultaneous equations e.g. x = 0 or x = -1.
Original post by davros
No it doesn't!

If one of the brackets is 0 then you just get 0 for the 1st term on the RHS leaving you with -3d + e.

You can plug in x = 1 to get a similar equation, or plug in lots of other small values of x to get a set of simultaneous equations e.g. x = 0 or x = -1.


aah i see what you guys are saying, my bad. Silly mistake on my part:biggrin:

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