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:'(

Printable View

- Jan 9th 2014, 06:05 PMgahyunkkPolynomials
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:'( - Jan 9th 2014, 06:12 PMchiroRe: Polynomials
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? - Jan 9th 2014, 06:18 PMgahyunkkRe: Polynomials
Its hsc math extension2 polynomial:)

Question is from cambridge 4units textbook.

Btw why did u take (x-r) out from the cubic equations where 'square root2 +r is rational and >0' ?

I still dont get how to solve this question fully :'( - Jan 9th 2014, 06:27 PMchiroRe: Polynomials
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. - Jan 9th 2014, 06:49 PMgahyunkkRe: Polynomials
Oh ok i see why u are taking r out from p(x)

Btw how do i actually solve the above question???:'(

Thank u so much for helping me :D - Jan 9th 2014, 07:36 PMHallsofIvyRe: Polynomials
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. - Jan 9th 2014, 07:50 PMgahyunkkRe: Polynomials
Oh why can -root2 be the other root?

I knew that if there is all real coefficient then p(x) has a complex conjugate zeros

But why -root2..?