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
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Guest

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

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

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

We must have $$2k^3- 9k+ 2= (k- 2)(2k^2+ 4k- 1)$$.

HallsofIvy

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