f(z) = z^3 - 2z + 4 , note how we type exponents
all complex roots , and irrational roots, come in conjugate pairs,
so the roots are 1+i and 1-i
sum of those is 2, product of those is also 2, so the quadratic producing
those is (z^2 - 2z + 2)
by long division that gives: (z+2)(z^2 - 2z + 2)
IF (z+2)(z^2 - 2z + 2) = 0
z = -2, 1 ± i
Consider the cubic polynomial
f(z) =z3−2z+ 4.
Given that z = 1 +i is a root of f(z), determine all the roots of f(z) in C
4 answers
How do you go about getting a quadratic from the sum and product? Sorry about this I'm not great with root questions
based on the fact that if you have ax^2 + bx + c = 0,
the sum of the roots = -b/a
and the product of the roots = c/a
often this is a avery efficient way to proceed, especially if the roots are irrational or complex.
Unfortunately, this does not appear to be on the curriculum of many regions, too bad.
e.g.
(2x+1)(x-3) = 0 , or 2x^2 - 5x - 3 = 0
from the factored form, x = -1/2 and 3
sum of those roots = -1/2 + 3 = 5/2
product of roots = (-1/2)(3) = -3/2
notice that -b/a = -(-5/2) = 5/2 and c/a = -3/2 , as I stated above
the sum of the roots = -b/a
and the product of the roots = c/a
often this is a avery efficient way to proceed, especially if the roots are irrational or complex.
Unfortunately, this does not appear to be on the curriculum of many regions, too bad.
e.g.
(2x+1)(x-3) = 0 , or 2x^2 - 5x - 3 = 0
from the factored form, x = -1/2 and 3
sum of those roots = -1/2 + 3 = 5/2
product of roots = (-1/2)(3) = -3/2
notice that -b/a = -(-5/2) = 5/2 and c/a = -3/2 , as I stated above
Okay that really helps. Thank you!
1 answer