Question 1183785
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Write an equation each of whose roots are 2 less than 3 times the roots of 3x^3 + 10x^2 + 7x - 10 = 0
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            This problem's solution is to apply the Vieta's theorem several times.



<pre>
Let  a,  b  and  c  be the roots of the given equation;

let  u,  v  and  w  be the roots of the projected equation.


According to the condition, we have 

    u = 3a-2,  v = 3b-2,  w = 3c-2.


    +----------------------------------------------------------------------+
    |   Let the projected equation be  px^3 + qx^2 + rx + s = 0.           |
    |                                                                      |
    |   Our goal is to determine the coefficients p, q, r and s.           |
    +----------------------------------------------------------------------+



According to Vieta's theorem,  a + b + c = {{{-10/3}}}.

Hence,  u + v + w = (3a-2) + (3b-2) + (3c-2) = 3*(a+b+c) - 6 = {{{3*(-10/3) - 6}}} = -10 - 6 = -16.

Thus    {{{q/p}}} = 16,  according to Vieta's theorem.



Next, according to Vieta's theorem,  ab + ac + bc = {{{7/3}}}.

Hence,  uv + uw + vw = (3a-2)*(3b-2) + (3a-2)*(3c-2) + (3b-2)*(3c-2) = 9(ab + ac + bc) - (6a + 6b + 6a + 6c + 6b + 6c) + (4+4+4) = 

                     = 9(ab + ac + bc) - 12(a+b+c) + 12 = {{{9*(7/3) - 12*(-10/3) + 12}}} = 21 + 40 + 12 = 73.

Thus    {{{r/p}}} = 73,  according to Vieta's theorem.



Finally, according to Vieta's theorem,  abc = {{10/3}}}.

Hence,  uvw = (3a-2)*(3b-2)*(3c-2) = 27abc - 18(ab + ac + bc) + 12(a + b + c) - 8 =  

                     = {{{27*(10/3) - 18*(7/3) + 12*(-10/3) - 8}}} = 9*10 - 6*7 - 4*10 - 8 = 0.

Thus    {{{s/p}}} = 0,  according to Vieta's theorem.



Since the projected polynomial coefficients ratios are integer numbers, we can take p = 1.

It gives q = 16,  r = 73,  s = 0.



Thus the projected equation  is  x^3 + 16x^2 + 73x = 0.      <U>ANSWER</U>
</pre>

Solved.


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<H3>The post-solution note</H3>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The solution by @MathLover1 works due to that happy occasion, &nbsp;that the given equation has a rational root.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;My approach / (my solution) &nbsp;works independently of this occasion.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;With my approach, &nbsp;there is no need in finding the roots of the given polynomial.