cubic polynomial

Divanhot SG

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

Divanhot SG

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

Divanhot SG

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

Divanhot SG

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

wee_ws SG

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!

Forbiddensinner SG

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.

wee_ws SG

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