Remainder theorem- Please help!

Remainder theorem- Please help!

Postby Guest » Fri Jan 03, 2020 6:51 am

Hi I have attached files of the question and solution.
Kindly advise the step for g(k)=(k-2)(2k^2+Bk-1) onwards until the end.
I do not understand the steps for the solution.
Thanks a lot!
Attachments
Solution_Q33.jpg
Solution_Q33.jpg (665.51 KiB) Viewed 105 times
Question_Q33.jpg
Question_Q33.jpg (250.34 KiB) Viewed 105 times
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Re: Remainder theorem- Please help!

Postby HallsofIvy » Fri Jan 17, 2020 9:32 am

You are given that [tex]f(x)= x^3+ kx^2- 7x+ 2[/tex] and are told that f(k)= 2k. Okay, [tex]f(k)= k^3+ k(k^2)- 7k+ 2= 2k[/tex]. Obviously [tex]k(k^2)= k^3[/tex] so that can be written as [tex]2k^3- 7k+ 2= 2k[/tex]. Subtracting 2k from both sides, [tex]2k^3- 9k+ 2= 0[/tex].

They then observe that [tex]2(2^3)- 9(2)+ 2= 16- 18+ 2= 0[/tex] so that, by the "remainder theorem", k- 2 is a factor of [tex]2k^3- 9k+ 2[/tex]. Further since [tex]2k^3- 9k+ 2[/tex] is of degree 3 and k- 2 is of degree 1, the other factor must be of degree 3- 1= 2: [tex]Ak^2+ Bk+ C[/tex] for some numbers A, B, and C. Multiplying, [tex](k- 2)(Ak^2+ Bk+ C)= Ak^3+ Bk^2+ Ck- 2Ak^2- 2Bk- 2C= Ak^3+ (B- 2A)k^2+ (C- 2B)k- 2C= 2k^2- 9k+ 2= 2k^3- 7k+ 2[/tex].

Comparing coefficients of "like powers", we must have [tex]A= 2[/tex], [tex]B- 2A= 0[/tex], [tex]C- 2B= -9[/tex], and [tex]-2C= 2[/tex]. The first and last of those give A= 2 and C= -1 immediately. Then the second and third are [tex]B- 4= 0[/tex], so B= 4, and [tex]C- 2B= -1- 2(4)= -9[/tex].

We must have [tex]2k^3- 9k+ 2= (k- 2)(2k^2+ 4k- 1)[/tex].

HallsofIvy
 
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