This is a trick question because if f(x) has a zero at 1,
that means f(1) = 0. So f(1) cannot equal -8. There can
therefore be no such polynomial function.
I will therefore pretend there was a typo and that the f(1)
should have been f(2) and that your question was this
legitimate question instead:
find the equation f(x) of degree three that has zeroes
at 1 and 1+i such that f(2)=-8?
The polynomial function that has zeros at r1, r2, r3, ..., rn
is of the form
f(x) = k(x-r1)(x-r2)···(x-rn)
You are given zeros 1 and 1+i
Therefore we have r1 = 1, r2 = 1+i, and r3 = 1-i
[The reason for r3 = 1-i is the fact that if a polynomial
with real coefficients has imaginary zero a+bi, it also has a zero
which is its conjugate a-bi.
So we have:
f(x) = k(x-1)[x-(1+i)]{x-(1-i)]
Remove the parentheses inside the brackets:
f(x) = k(x-1)[x-1-i]{x-1+i]
Put parentheses around the (x-1)'s:
f(x) = k(x-1)[(x-1)-i][(x-1)+i]
Multiply the bracketed expressions using FOIL:
f(x) = k(x-1)[(x-1)²+i(x-1)-i(x-1)-i²]
The two middle terms cancel:
f(x) = k(x-1)[(x-1)²-i²]
Replace i² by (-1)
f(x) = k(x-1)[(x-1)²-(-1)]
f(x) = k(x-1)[(x-1)²+1]
f(x) = k(x-1)[x²-2x+1+1]
f(x) = k(x-1)(x²-2x+2)
f(x) = k(x³-2x²+2x-x²+2x-2)
f(x) = k(x³-2x²+2x-x²+2x-2)
f(x) = k(x³-3x²+4x-2)
Now since we are given that f(2) = -8,
we substitute x=2
f(2) = k(2³-3·2²+4·2-2)
We substitute -8 for f(2)
-8 = k(8-3·4+8-2)
-8 = k(8-12+6)
-8 = k(2)
-8 = 2k
-4 = k
Substitute -4 for k in
f(x) = k(x³-3x²+4x-2)
f(x) = -4(x³-3x²+4x-2)
f(x) = -4x³+12x²-16x+8
Edwin
