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# Easy remainder theorem problem Watch

1. 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!!
2. (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 that way, and if you substitute you'll get another expression which is . 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.
3. (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!
4. (Original post by Smaug123)
You could indeed find out that way, and if you substitute you'll get another expression which is . 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.
5. (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?
6. (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.
7. (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
8. (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

Thanks again!!
9. (Original post by brianeverit)
Quite correct.
Thank you!!
10. (Original post by Smaug123)
You could indeed find out that way, and if you substitute you'll get another expression which is . 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!!
11. (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?
12. (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.
13. (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.
in my humble opinion what you are being told is correct but certainly a quick long division is the most effective method
14. (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!
15. (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)
16. (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!!
is 0. That entire bracket vanishes.
17. (Original post by Smaug123)
is 0. That entire bracket vanishes.
ok. Which leaves me with 9 - 3b + c -3d + e
18. (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

(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
19. (Original post by JamesNeedHelp2)
ok. Which leaves me with 9 - 3b + c -3d + e
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.
20. (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

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