Thread: 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

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

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.

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

If the polynomial is even, then for every root you have another root (of the same order). Since you have only two roots left, and their product must be -2, they must have absolute value ; and they must be real, because if they were a pair of conjugate complex numbers their product would be .