Polynomial Question?

Question

The cubic polynomial f(x) is such that the coefficient of #x^3# is -1 and the roots of the equation f(x)=0 are 1, 2 and k. Given that f(x) has a remainder of 8 when divided by x-3, find:
i. The value of k
ii. The remainder when f(x) is divided by x+3

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Answer 1 · 2018-08-04T15:42:17

#k=7# and the remainder when divided by #x+3# is #200#

Given that #f(x)# has zeros #1, 2, k#, it has factors #(x-1)#, #(x-2)# and #(x-k)#.

So with the additional information that the coefficient of the leading term is #-1#, we can write:

#f(x) = -(x-1)(x-2)(x-k)#

The remainder #8# when divided by #(x-3)# is the value of #f(3)#, so:

#8 = f(3)#

#color(white)(8) = -((color(blue)(3))-1)((color(blue)(3))-2)((color(blue)(3))-k)#

#color(white)(8) = 2k-6#

Hence #k=(8+6)/2 = 7#

Then the remainder when divided by #(x+3)# is #f(-3)#:

#f(-3) = -((color(blue)(-3))-1)((color(blue)(-3))-2)((color(blue)(-3))-7)#

#color(white)(f(-3)) = 200#

graph{(y+(x-1)(x-2)(x-7))=0 [-4, 10, -15, 32]}