The cubic polynomial f(x) is such that the coefficient of x^3 is -2 and the roots of the equation f(x)=0 are 1,3 and k. Given that f(x) has a remainder of 16 when divided by x-5, find:
(i) the value of k
(ii) an expression for f(x) in descending powers of x.
(iii) the remainder when f(x) is divided by 2x+1
my calculation for part (i)
let f(x)= (x-1)(x-3)(x-k)
f(5)=16, sub x=5 into f(x), k=3
am i correct?
for part (ii) i am stuck
or is answer to part (i)
f(x)= -2(x-1)(x-3)(x-k)
f(5)=16, , subst x=6 into f(x), k=6
can someone verify part (ii) answer is f(x)= -2(x-1)(x-3)(x-6)= -2(x^3-10x^2+27x-18)
for part (iii) ans, remainder =68.25
thxs
Originally posted by Divanhot:or is answer to part (i)
f(x)= -2(x-1)(x-3)(x-k)
f(5)=16, , subst x=6 into f(x), k=6
Correct, good!
Originally posted by Divanhot:or is answer to part (i)
f(x)= -2(x-1)(x-3)(x-k)
f(5)=16, , subst x=6 into f(x), k=6
This is the answer to part (i).
For (ii), the answer you have given should be correct ( Note that I am in a rush to somewhere else, so I did not check your workings. ), but the -2 must be put in.
For (iii), if you sub in x= -1/2, the answer should be correct.
Originally posted by Divanhot:can someone verify part (ii) answer is f(x)= -2(x-1)(x-3)(x-6)= -2(x^3-10x^2+27x-18)
for part (iii) ans, remainder =68.25
thxs
Correct again, great effort!
Cheers,
Wen Shih