Expert: Socrates - 9/18/2011

Question
1.find the polynomial function of least degree with integral coefficient which has 2-squareroot of 3 and 1-i among its zeros.
2.find the polynomial function p(x) if it is of the second degree and p(1)=3, p(0)=-1, and p(-1)=-3.
3.if f(x)=ax2+bx+c and f(x+3)=X2+7x+4, what is a, b, c?

Answer
1)

Since complex zeros for polynomials with real coefficients always occur in conjugate pairs, 1+i must also be a zero.

Thus , 2√3 , 1+i , 1-i must all be zeros.

Note that (x-(1+i))(x-(1-i)) = x^2-2x+2 , and so x^2-2x+2 is a factor of the polynomial we want.

2√3 is a root of x^2-12 , and x^2-12 doesn't factor, so x^2-12 must also be a factor of the polynomial we want


This gives us (x^2-2x+2)(x^2-12) = x^4 - 2x^3 - 10x^2 + 24x - 24 for the polynomial with least degree


2)

Let the polynomial be p(x)= ax^2 + bx + c

3 = p(1) = a+b+c

-1 = p(0) = c

-3 = p(-1) = a-b+c


solving for a , b , c gives a = 1 ,  b = 3 ,  c = -1


The answer is  p(x) = x^2 + 3x - 1



3)

substitute x+3 in for x in ax^2+bx+c and get f(x+3) =  ax^2 + (6a+b)x + 9a+3b+c

We also know f(x+3) =  x^2+7x+4  , so  a = 1 , 6a+b = 7 and 9a+3b+c = 4


Solve for a , b, c and get

a=1 , b=1 , c=-8