Question 46190
<pre><font size = 4><b>Find the polynomial f(x) of degree 
three that has zeroes at 1.2 and 4 such that f(0) = -16.

<i><font color = "red">Unlike the previous tutor, I'm assuming " 1.2 " was a typo,
and you really meant the "." to be a comma rather than a decimal.
This, in other words:</i></font>

<b>Find the polynomial f(x) of degree three that has zeroes 
at 1, 2 and 4 such that f(0) = -16.

All polynomials of degree n that have zeros r<sub>1</sub>, r<sub>2</sub>, ..., r<sub>n</sub> 
are of the form

f(x) = k(x-r<sub>1</sub>)(x-r<sub>2</sub>)···(x-r<sub>n</sub>) 

where k can be any non-zero number.

So any polynomial of degree 3 that has zeros 1, 2, and 4 is 
of the form

f(x) = k(x-1)(x-2)(x-4)

where k can be any non-zero number.

But we also want f(0) to equal -16. So it has to be true that 
if we substitute 0 for x and then -16 for f(0), they should be 
equal.  So,

f(x) = k(x-1)(x-2)(x-4)
f(0) = k(0-1)(0-2)(0-4)
 -16 = -8k
   2 = k

So substitute 2 for k in 

f(x) = k(x-1)(x-2)(x-4)

to give

f(x) = 2(x-1)(x-2)(x-4)    

Multiply all that out and you'll get

f(x) = 2x³ - 14x² + 28x - 16 

Edwin</pre>