Question 16789: I'm problems with the following problem in understanding it.
Find a polynomial with leading coefficient 1 and degree 3 that has -1, 1, and 3 as roots.
It seems pretty easy and I think the answer is x^3-3x^2-x-3, but that just from looking at it. I'm not sure how to actually do the problem.
Thanks for any help
Answer by xcentaur(357)   (Show Source):
You can put this solution on YOUR website!
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
x^3-3x^2-x+3=0
Now this can be verified for a polynomial as:
(x-p)(x-q)(x-r)=f(x)
(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.
Hope this helps,
Prabhat
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