# Factorise a polynomial without given the equation

• April 12th 2010, 08:28 PM
differentiate
Factorise a polynomial without given the equation
P(x) is an even monic polynomial of degree 4 with integer coefficients. One zero is 2i and the product of the zeros is -8. Factorise P(x) over R.

the answer is -2i, 2i, root 2, - root2

i know that if 2i is a root, then -2i is also another root, but I dont get how they got root 2 and - root 2

I can let the other roots be "a" and "b"
where ab = -2 (since product of roots is -8), however I don't understand how you could get root 2, - root 2 from that

Thanks (Rofl)
• April 12th 2010, 08:42 PM
sa-ri-ga-ma
Quote:

Originally Posted by differentiate
P(x) is an even monic polynomial of degree 4 with integer coefficients. One zero is 2i and the product of the zeros is -8. Factorise P(x) over R.

the answer is -2i, 2i, root 2, - root2

i know that if 2i is a root, then -2i is also another root, but I dont get how they got root 2 and - root 2

I can let the other roots be "a" and "b"
where ab = -2 (since product of roots is -8), however I don't understand how you could get root 2, - root 2 from that

Thanks (Rofl)

I can let the other roots be "a" and "b"
Then the product of roots is 2i*-2i*ab = -8 . So ab = -2.
In an even monic polynomial odd powered x terms are missing.
So the sum of the roots must be zero.
So a+b = 0 and ab = -2. Now find a and b.
• April 12th 2010, 09:01 PM
Failure
Quote:

Originally Posted by differentiate
P(x) is an even monic polynomial of degree 4 with integer coefficients. One zero is 2i and the product of the zeros is -8. Factorise P(x) over R.

the answer is -2i, 2i, root 2, - root2

i know that if 2i is a root, then -2i is also another root, but I dont get how they got root 2 and - root 2

I can let the other roots be "a" and "b"
where ab = -2 (since product of roots is -8), however I don't understand how you could get root 2, - root 2 from that

Thanks (Rofl)

If the polynomial is even, then for every root $a\neq 0$ you have another root $b=-a$ (of the same order). Since you have only two roots left, and their product must be -2, they must have absolute value $\sqrt{2}$; and they must be real, because if they were a pair of conjugate complex numbers $x\pm i\cdot y$ their product would be $x^2+y^2>0$.