# find the cubic polynomial

#### 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?

#### simplependulum

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

Then sub. back to the equation given :

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

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

Squaring gives :

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

Try to do the second one by this method .

• TheCoffeeMachine

#### TheCoffeeMachine

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 $$\displaystyle x_{1}+x_{2}+x_{3} = -7$$, $$\displaystyle x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2} = -8[/Math], and \(\displaystyle x_{1}x_{2}x_{3} = -3$$. So $$\displaystyle 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$$\)$$\displaystyle and \(\displaystyle 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})$$ $$\displaystyle = (-8)^2-2(-3)(-7) = 22$$ and $$\displaystyle x_{1}^2x_{2}^2x_{3}^2 = (x_{1}x_{2}x_{3})^2 = (-3)^2 = 9$$, which gives $$\displaystyle x^3+65x^2+22x+9$$. The signs alternate, so the required equation is $$\displaystyle x^3-65 x^2+22 x-9$$, which simplependulum's equality yields as well.\)

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