Given roots, write the equation...

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I can't get my head around this, please help me out!

Directions: Determine the least degree polynomial function with integral coefficients that gives the following roots:

1/2, 2

In class today we learned to solve these problems using sums and products of roots. I tried:

sum of roots: 2.5
product of roots: 1

x^2 - (sum) x+ (product) = 0

x^2 -2.5x + 1 = 0

...but -2.5 isn't integral. :/

I realize the root 1/2 factors to (x-1/2) but also (2x-1), so the answer is probably:

(2x-1)(x-2) = 2x^2 - 5x + 2

...but how do I achieve that with the sums/roots method? There are many other (more difficult) problems on my homework with fractions, so it's important I can figure this out.

Thanks a bunch!
 
One could just write:
(x−1/2)(x− 2)=0...............This is the factor/root theorem.
(2x−1)(x− 2)=0
2x<SUP>2</SUP>−5x+2=0.
That is what you got!
 
ok, but would it be possible to use the same method on a three root problem?

eg:

-1/2, 3, -2

How would I do that?
 
sum of roots: 2.5
product of roots: 1


x + y = 2.5
xy = 1
x + 1/x = 2.5 or x^2 - 2.5 + 1 = 0
From the quadratic equation, x = [2.5+/-sqrt(6.25 - 4)]/2
x = 2 and 1/2.

x + y = 2.5 and xy = 1
 
TchrWill said:
sum of roots: 2.5
product of roots: 1


x + y = 2.5
xy = 1
x + 1/x = 2.5 or x^2 - 2.5 + 1 = 0
From the quadratic equation, x = [2.5+/-sqrt(6.25 - 4)]/2
x = 2 and 1/2.

x + y = 2.5 and xy = 1
Click to expand...

but 2.5 has to be an integer, I think
 
“same method on a three root problem? eg: -1/2, 3, -2"
Yes indeed!

(x+1/2)(x−3)(x+2)=0

(2x+1)(x−3)(x+2)=0

Multiply that out.
 
Whee! Thank you :)

Okay. Other thing. The only root given is 2i, how would I do that one?
 
"The only root given is 2i, how would I do that one?"
In the real field complex roots occur is conjugate pairs.
That is if a+bi is a root then so is a−bi.
Thus, if 2i is a root the so is −2i, or the factors are (x−2i) & (x+2i)
 
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