P(x) is a monic polynomial of degree 4 with integer coefficients and constant term 4. One zero is root2, another zero is rational and the sum of the zeros is positive. Factorise p(x) fully over Real field.
Help me:'(
Hey gahyunkk.
For these types of problems, its best to translate your given information into mathematical expressions/equations. Given:
P(x) = ax^4 + bx^3 + cx^2 + dx + 4 with a,b,c,d integer values we also have
P(x) = (x - SQRT(2))*(fx^3 + gx^2 + hx + k)
= (x - SQRT(2)(x - r)(jx^2 + ox + p)
where SQRT(2) + r is rational and > 0.
You can also use other results like the rational root theorem for your problem.
What course is this for by the way?
Basically if r is a root of P(x) then we can factor P(x) in that way. The reason why is due to the following:
If P(r) = 0 then this implies that P(x) can be factored into P(x) = (x-r)Q(x) where Q(x) is of degree n - 1 if P(x) is of degree n. for n > 1. This is from result of polynomial factoring where
P(x) = D(x)Q(x) + R(x) or in other words, if you calculate P(x)/D(x), you get Q(x) plus some remainder R(x). If you have a root, then R(x) = 0. This is why you can factor as I did above.
It looks to me like there are an infinite number of different solutions. Since one zero is $\displaystyle \sqrt{2}$ and the coefficients are integer, we are going to have to take another to be $\displaystyle -\sqrt{2}$. There are, then, two other zeros, call them "a" and "b", with at least a rational. The polynomial must be of the form $\displaystyle c(x- \sqrt{2})(x+ \sqrt{2})(x- a)(x- b)= c(x^2- 2)(x- a)(x- b)$. The constant term is $\displaystyle c(-\sqrt{2})(\sqrt{2})(-a)(-b)= -2abc= 4$. The sum of the zeros is $\displaystyle \sqrt{2}- \sqrt{2}+ a+ b= a+ b$ and that must positive.
So any a, b, c such that a is rational, abc= -2, and a+b> 0 will work.