Polynomial Help (1 Viewer)

[nimrod_dookie]

Polynomial P(x)=x^4+ax^3+bx^2+cx-10 with real coeff, has 2 integer zeros p and q.

If p(x) has complex zero k-i where k is an integer.
Use this zero to obtain a real quadratic factor of p(x)
state all possible values of k for which this quadratic is a factor of p(x)

Using p and q write another expression for a real quadratic of p(x). hence list all possible values of pq for which p,q and k-i are zeros of p(x).

Given p+q=-9 show that there is only one possible value for pq.
Hence, find the polynomial p(x) in expanded form and all zeros of p(x)

If anyone could offer assistance it would be greatly appreciated.

 

[Riviet]

Given k-i is a root, then k+i must also be a root by complex conjugate root theorem.

Then a quadratic factor of p(x) is [x-(k-i)][x-(k+i)] = x²-2kx+k²+1.

Also the other quadratic factor of p(x) would be (x-p)(x-q) = x²-(p+q)x+pq.

 

[pesila]

Okay I tried to post a nice one earlier but it failed, so here goes a little more rushed one. k-i is a factor, so as is k+i by conjugate dealies. p and q are factors. the product equation is (k-i)(k+i)pq = -10. pq(k^2 +1) = -10. Since pq is the product of two integers and k is an integer, k^2 +1 can equal 1, 2, 5, 10 for k = 0,±1,±2,±3. and pq = -10, -5, -2, -1 respectively. The quadratic with p and q is x^2 - (p+q) +pq = 0. with p+q = -9, x^2 + 9x +pq = 0. Trying values for pq as stated above, we find that only pq = -10 gives integers as its solutions. Therefore p and q are -10 and 1. From the product of the quartic, -10(k^2 +1) = -10. Therefore k=0. This makes the roots -10, 1, i, -i. The quartic becomes (x+10)(x-1)(x-i)(x+i)=0. By expansion I got x^4 + 9x^3 - 9x^2 + 9x - 10 = 0. Hope that helps. It's always hard to explain these things.