# Math Help - Polynomial question

1. ## Polynomial question

Hey guys
I'm stuck on this polynomial problem, i think its pretty simple though I just cant do it lol, basically trying to factorise so that there is no surd left

Heres the question:

If $\alpha,\beta,\gamma,\delta$ are the roots of $x^{4}-x^{2}+2x+3=0$, form the equation whose roots are: $\alpha^{2},\beta^{2},\gamma^{2},\delta^{2}$

So this is what I did

Let $y=x^{2}$
$x=y^\frac{1}{2}$
substituting into equation

$(y^\frac{1}{2})^{4}-(y^\frac{1}{2})^{2}+2(y^\frac{1}{2})+3=0$

$y^{2}-y+2(y^\frac{1}{2})+3=0$

yea this i where I've tried to manipulate the equation so to obtain a polynomial with whole number powers...but I haven't been able to do it.. looks simple enough

2. Originally Posted by aonin
Hey guys
I'm stuck on this polynomial problem, i think its pretty simple though I just cant do it lol, basically trying to factorise so that there is no surd left

Heres the question:

If $\alpha,\beta,\gamma,\delta$ are the roots of $x^{4}-x^{2}+2x+3=0$, form the equation whose roots are: $\alpha^{2},\beta^{2},\gamma^{2},\delta^{2}$

So this is what I did

Let $y=x^{2}$
$x=y^\frac{1}{2}$
substituting into equation

$(y^\frac{1}{2})^{4}-(y^\frac{1}{2})^{2}+2(y^\frac{1}{2})+3=0$

$y^{2}-y+2(y^\frac{1}{2})+3=0$

yea this i where I've tried to manipulate the equation so to obtain a polynomial with whole number powers...but I haven't been able to do it.. looks simple enough

.

3. Originally Posted by aonin
Hey guys
I'm stuck on this polynomial problem, i think its pretty simple though I just cant do it lol, basically trying to factorise so that there is no surd left

Heres the question:

If $\alpha,\beta,\gamma,\delta$ are the roots of $x^{4}-x^{2}+2x+3=0$, form the equation whose roots are: $\alpha^{2},\beta^{2},\gamma^{2},\delta^{2}$

So this is what I did

Let $y=x^{2}$
$x=y^\frac{1}{2}$
substituting into equation

$(y^\frac{1}{2})^{4}-(y^\frac{1}{2})^{2}+2(y^\frac{1}{2})+3=0$

$y^{2}-y+2(y^\frac{1}{2})+3=0$

yea this i where I've tried to manipulate the equation so to obtain a polynomial with whole number powers...but I haven't been able to do it.. looks simple enough

Have you already studied Viete's formulae? Otherwise google it. According to these, as $\alpha\beta\gamma\delta=3$ ,we then

get that $\alpha^2\beta^2\gamma^2\delta^2=9$. We get as well:

$\alpha+\beta+\gamma+\delta=0\Longrightarrow \alpha^2+\beta^2+\gamma^2+\delta^2=(\alpha+\beta+\ gamma+\delta)^2-2(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamm a+\beta\delta+\gamma\delta)=$

$=0-2(-1)=2$ , and etc. In this way you get your quartic polynomial's coefficients

Tonio

4. Hi, tonio thanks for the tip. I'm aware of it, although I'm trying to solve it using the method that my textbook uses.this is an example of the question using the factorisation and squaring of both sides: seems pretty simple aint it? except i can't seem to work it out for the one im doing

EXAMPLE QUESTION

If $\alpha,\beta,\gamma,\delta$ are the roots of $x^{3}+2x^{2}-2x+3=0$, form the equation whose roots are: $\alpha^{2},\beta^{2},\gamma^{2},\delta^{2}$

$Let$
$y=x^{2}$

$x=y^\frac{1}{2}$

$(y^\frac{1}{2})^{3}+2(y^\frac{1}{2})^{2}-2(y^\frac{1}{2})+3=0$

$y^\frac{3}{2}+2y-2y^\frac{1}{2}+3=0$

$y^\frac{3}{2}-2y^\frac{1}{2}=-3-2y$

$y^\frac{1}{2}(y-2)=-3-2y$

square both sides to get

$y(y-2)^{2}=9+12y+4y^{2}$

$y^{3}-4y^{2}+4y=9+12y+4y^{2}$

$y^{3}-8y^{2}-8y-9=0$

I was wondering with my original question..would i be able to factorise out so that squaring both sides would eliminate a surd

5. Originally Posted by aonin
Hi, tonio thanks for the tip. I'm aware of it, although I'm trying to solve it using the method that my textbook uses.this is an example of the question using the factorisation and squaring of both sides: seems pretty simple aint it? except i can't seem to work it out for the one im doing

EXAMPLE QUESTION

If $\alpha,\beta,\gamma,\delta$ are the roots of $x^{3}+2x^{2}-2x+3=0$, form the equation whose roots are: $\alpha^{2},\beta^{2},\gamma^{2},\delta^{2}$

$Let$
$y=x^{2}$

$x=y^\frac{1}{2}$

$(y^\frac{1}{2})^{3}+2(y^\frac{1}{2})^{2}-2(y^\frac{1}{2})+3=0$

$y^\frac{3}{2}+2y-2y^\frac{1}{2}+3=0$

$y^\frac{3}{2}-2y^\frac{1}{2}=-3-2y$

$y^\frac{1}{2}(y-2)=-3-2y$

square both sides to get

$y(y-2)^{2}=9+12y+4y^{2}$

$y^{3}-4y^{2}+4y=9+12y+4y^{2}$

$y^{3}-8y^{2}-8y-9=0$

I was wondering with my original question..would i be able to factorise out so that squaring both sides would eliminate a surd

$y^2-y+2y^{1/2}+3=0\Longrightarrow 2y^{1/2}=-(y^2-y+3)\Longrightarrow 4y=y^4-2y^3+7y^2-6y+9$ , and there