# Thread: find the cubic polynomial

1. ## find the cubic polynomial

Let x₁, x₂, x₃ be the roots of f(x)=x^3+7x^2-8x+3.
Find the cubic polynomial whose roots are
a) x₁^2, x₂^2, x₃^2
b) 1/(x₁), 1/(x₂), 1/(x₃)
Because we are learning resolvent cubic now, so I think that this may relate to it. Can someone tell me how to find them?

2. Originally Posted by apple2009
Let x₁, x₂, x₃ be the roots of f(x)=x^3+7x^2-8x+3.
Find the cubic polynomial whose roots are
a) x₁^2, x₂^2, x₃^2
b) 1/(x₁), 1/(x₂), 1/(x₃)
Because we are learning resolvent cubic now, so I think that this may relate to it. Can someone tell me how to find them?
I tell you the technique

Let $y = x_1^2$ as a root of the new polynomial we are looking for . The key is to write the original root as the subject . ie. $x_1 = \sqrt{y}$

Then sub. back to the equation given :

$\sqrt{y}^3 + 7y - 8\sqrt{y} + 3 = 0$

$\sqrt{y} ( y-8) = -(7y+3)$

Squaring gives :

$y(y-8)^2 = (7y+3)^2$ I leave the expansion to you .

Try to do the second one by this method .

3. Originally Posted by apple2009
Let x₁, x₂, x₃ be the roots of f(x)=x^3+7x^2-8x+3. Find the cubic polynomial whose roots are x₁^2, x₂^2, x₃^2
Using Vieta's relations, we see that $x_{1}+x_{2}+x_{3} = -7$, $x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2} = -8[/tex], and img.top {vertical-align:15%;} $x_{1}x_{2}x_{3} = -3$. So $$x_{1}^{2}+x_{2}^{2}+x_{3}^2= (x_{1}+x_{2}+x_{3})^2-2\left(x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2}\right) = (-7)^2-2(-8) = 65 " alt="x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2} = -8$$, and $x_{1}x_{2}x_{3} = -3$. So [tex] x_{1}^{2}+x_{2}^{2}+x_{3}^2= (x_{1}+x_{2}+x_{3})^2-2\left(x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2}\right) = (-7)^2-2(-8) = 65
" /> and $x_{1}^2x_{2}^2+x_{1}^2x_{3}^2+x_{3}^2x_{2}^2 = (x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2})^2-2(x_{1}x_{2}x_{3})(x_{1}+x_{2}+x_{3})$ $= (-8)^2-2(-3)(-7) = 22$ and $x_{1}^2x_{2}^2x_{3}^2 = (x_{1}x_{2}x_{3})^2 = (-3)^2 = 9$, which gives $x^3+65x^2+22x+9$. The signs alternate, so the required equation is [LaTeX ERROR: Convert failed] , which simplependulum's equality yields as well.