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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 \(2r_3\)

 May 26, 2019
 #1
avatar+107348 
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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

 

 

cool cool cool

 May 26, 2019
edited by CPhill  May 27, 2019

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