Write an equation whose roots are the roots of the equation x^3+6x^2-5x+7=0 each multiplied by 2?

Answer 1

x^3 + 6x^2 - 5x +7 = 0

This may be written as (x - a)(x - b)(x - c) = 0,
where a, b, c are the roots.

Expanding gives :
x^3 - (a + b + c)x^2 + (ab + bc + ca)x - abc = 0

Equating coefficients of this equation with the original gives :
a + b + c = -6, ab + bc + ca = -5 and abc = - 7.

If the roots a, b, c are each multiplied by 2, then the
final equation will be : (x - 2a)(x - 2b)(x - 2c) = 0

Expanding gives :
x^3 - 2(a + b + c)x^2 + 4(ab + bc + ca)x - 8abc = 0

Substituting in the values for
a + b + c, ab + bc + ca and abc gives :
x^3 - 2(-6)x^2 + 4(-5)x - 8(-7) = 0

which finally gives :
x^3 + 12x^2 - 20x + 56 = 0

Answer 2

An easier way would be to just replace x with x/2 in the equation.

In other words let y = 2x then substitute x = y/2 in the equation. which amounts to the same thing.

This gives (y/2)^3+6(y/2)^2-5(y/2)+7 = 0
(y^3)/8 + (6y^2)/4 - (5y)/2 + 7 = 0
multiplying by 8 gives:
y^3 + 12y^2 - 20y + 56 = 0

Since y = 2x, if x is a solution to the original equation, y = 2x will be a solution of the new equation.