Question 16789
<pre>
Now let the polynomial be 

x^3-ax^2+bx-c=0

Let the roots be p,q,and r.

Then a=p+q+r
     b=pq+qr+pr
     c=pqr

since the roots are given as -1,1 and 3,we get

a=-1+1+3=3

b=(-1*1)+(1*3)+(-1*3)=-1

c=(-1)(1)(3)=-3

substituting,
x^3-ax^2+bx-c=0

x^3-(3)x^2+(-1)x-(-3)=0

<b><font size=+2>x^3-3x^2-x+3=0</font></b>
<br>
Now this can be verified for a polynomial as:

(x-p)(x-q)(x-r)=f(x)
<br>
(x-(-1))(x-(+1))(x-(+3))
(x-1)(x+1)(x-3)
On expanding we get the expression that we just worked out. Thus we know that our polynomial is indeed correct.
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Hope this helps,
Prabhat