Find values of k so that each remainder is 3
\[(x^2+kx-17)\div(x-2)\] the key is 8
1 k -17 2 2 4+2k ______________ 1 2+k 13+2k
k=10
do you know synthetic division because that is what i used. the remainder is 13+2k so set \[13+2k=3\] and solve for k via \[13+2k=3\] \[2k=-10\] \[k=-5\] but my algebra could be wrong.
yeah it is wrong. remainder should be -13+2k \[-13+2k=3\] \[2k=26\] \[k=13\]
1 k -17 2 2 4+2k ______________ 1 2+k 13+2k how u get 4+2k
i multiplied 2 times 2+2k but i think i made a mistake somewhere. in any case the remainder should be -13+2k not 13+2k
let me do it on paper and get it right.
ok now i have it. k = 8
first off is synthetic division clear? i mean do you know what i am doing when i write 1 k -17 2 __________________________
i am dividing using synthetic division. it is pretty easy. 1 k -17 2 2 4+2k __________________________ 1 2+k -13+k
x^2 + kx -17 2 1 1 -17 ----------2----4-------- 1 2 -13 ?
you forgot that when you get the 2 you have to add 2 + k
you do not know what the "k" is so when you multiply 1 by 2 you get 2, put it in the line under the k and then add to get 2+k then you multiply 2 + k by 2, get 4+2k and put it under the -17. when you add you get -13+2k as your remainder
the coefficient of the "x" term is k, not 1
so the set up for the divisions should look like 1 k -17 2 __________________________
1 k -17 2 frist line 1 from what number, k from what number? I know second line divi by 2
you list the coefficients of \[x^2+kx-17\] which are 1 k -17
Ok I got it , Thank
ok good. if you don't know synthetic division this would be a pain. so you get a remainder of -13+2k which you set = 3 and solve to get k = 8. that is the correct answer and if you divide \[\frac{x^2+8x-17}{x-2}\] you will get \[x+10+\frac{3}{x-2}\] i.e. \[(x+10)(x-2)+3=x^2+8x-17\]
I kow very well synthetic division ,I just don't understand how divi without don't know k , now I get it , ty
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