Dividing polynomials
x3+3x+k, put this into the form (ax2+bx+c)(x-1) and show k=2, the remainder is 6
First I got the form (ax2+bx+c)(x-1)
multiplied ax3+bx2+cx-ax2-bx-c
grouped like terms ax3+(b-a)x2+(c-b)x-c
Put original equation into that form 1x3+0x2+3x-(-k)
1 is a, b-1=0 therefore b is also 1, c is -k, -k-1=3 therefore k should be 4 but its not, the answers (which do not show the working out) says its 2, which does satisfy the original equation if I substitute x=1.
First I got the form (ax2+bx+c)(x-1)
multiplied ax3+bx2+cx-ax2-bx-c
grouped like terms ax3+(b-a)x2+(c-b)x-c
Put original equation into that form 1x3+0x2+3x-(-k)
1 is a, b-1=0 therefore b is also 1, c is -k, -k-1=3 therefore k should be 4 but its not, the answers (which do not show the working out) says its 2, which does satisfy the original equation if I substitute x=1.
(edited 11 years ago)
Original post by Primus2x
Oh sorry it says the remainder is 6
So you have not given anything like the original question ... why not?
Anyway 1+3+k=6 gives k=2 as required
Original post by Primus2x
ax3+(b-a)x2+(c-b)x-c
ax3+(b-a)x2+(c-b)x-c
With your new information this will read
So this method also gives c=4 and k=2
Original post by Primus2x
I see it makes sense now so if f(x) gives a remainder when divided by (x-n) it means it is equal to the remainder when I substitute x=n into it?
That is indeed the Remainder Theorem
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