Find a polynomial whose zeroes are double the zeroes of P(x)=x^3-7x^2-5x+2.

Answer 1

Properties of the roots of a cubic
If the roots of a cubic are a, b , and c
then start with (x-a)(x-b)(x-c)=0 and multiply out.
We get x3-(a+b+c)x2+(ab+bc+ca)x-(abc)=0

sum zeros is [-(coeffiecient of x^2 term) ]
sum of products of two at a time = coeffi
so for your equation we get

a+b+c = 7
ab + bc + ac = -5
abc = -2

let the zeros of our new equation be
2a, 2b and 2c
2a+2b+2c =
2(a+b+c) = 2(7) = 14
4ab + 4bc + 4ac = 4(ab + bc + ac) = 4(-5) = -20
(2a)(2b)(2c)
= 8abc = 8(-2) = -16

so without finding the actual roots we have
f(x) = x^3-14x^2 - 20x - 16

Answer 2

my second paragraph cut a bit mangled, has no effect on the rest.

should have been:

sum zeros is [-(coeffiecient of x^2 term) ]
sum of products of two at a time = coefficient of x term
product of all three roots = -constant term

so for your equation we get
etc (rest is good)