+170 Let \(r_1\) \(, r_2\) and \(r_3\) be the roots of \(x^3 - 3x^2 + 8 = 0.\)
Find the monic polynomial, in \(x\) whose roots are \(2r_1, 2r_2\) and \(2r_3\)
+128408 x^3 - 3x^2 + 8 = 0
We have the form
ax^3 - bx^2+ cx + d = 0
Using Vieta's Theorem, we have that
The sum of the roots = -( -b)/a = r1 + r2 + r3 = --(-3)/1 = 3
And the product of the roots r1 * r2 * r3 = -d/a = -8/1 = - 8
So....the sum of the new roots
2r1 + 2r2 + 2r3 = 2 (r1 + r2 + r3) = 2(3) = 6 ⇒ -b/1 ⇒ b = -6
And the product of the new roots = 2r1 * 2r2 * 2r3 = 8(r1*r2*r3) = 8(-8) = -64 ⇒ -d / 1 ⇒ d = 64
So the new polynomial is
x^2 - 6x^2 + 64
