Complex numbers (1 Viewer)

yeyjasminetime · Oct 12, 2013

Hi,

How do you solve for x?

X^3 + x +2

Thanks

Sy123 · Oct 12, 2013

yeyjasminetime said:
Hi,

How do you solve for x?

X^3 + x +2

Thanks

$x^3+x+2 = (x^3+1) + (x+1) = (x+1)(x^2-x+1) + (x+1)$

I'll let you do the rest

----

Alternatively, the easier method:

We do the 'sub in values', method, where you sub in values that are factors of 2, so sub in, x=-1, x=1, x=2, x=-2, see which one gives a solution.
We find that x=-1 is a solution
Therefore (x+1) is a factor

$\therefore \ \ x^3+x+2 = (x+1)(ax^2 + bx +c)$

Then equate co-efficients to find a, b, and c then find using quadratic formula or otherwise to find the roots of the quadratic

yeyjasminetime · Oct 12, 2013

Could you explain the harder method?

Sy123 · Oct 12, 2013

yeyjasminetime said:
Could you explain the harder method?

The 'harder' method comes mostly from intuition, I wanted to manipulate the expression in a way that I can factorize something out, and I noticed that the 2 can be split up.
Don't worry about that method though, the x=-1 substitution one is much more direct

SpiralFlex · Oct 12, 2013

Another method to solve such cubics is Cardano's method

QZP · Oct 12, 2013

Sy123 said:
$x^3+x+2 = (x^3+1) + (x+1) = (x+1)(x^2-x+1) + (x+1)$

I'll let you do the rest

----

Alternatively, the easier method:

We do the 'sub in values', method, where you sub in values that are factors of 2, so sub in, x=-1, x=1, x=2, x=-2, see which one gives a solution.
We find that x=-1 is a solution
Therefore (x+1) is a factor

$\therefore \ \ x^3+x+2 = (x+1)(ax^2 + bx +c)$

Then equate co-efficients to find a, b, and c then find using quadratic formula or otherwise to find the roots of the quadratic

I don't understand the second method. Factors of 2? How/why? :S

SpiralFlex · Oct 12, 2013

I wrote up a proof but I accidentally deleted it (on phone- closed the thread)

SpiralFlex · Oct 12, 2013

QZP said:
I don't understand the second method. Factors of 2? How/why? :S

$$Suppose we have some polynomial equation$ \; a_{n}x^n + a_{n-1}x^{n-1}+...+a_{0}=0 \; $given that each rational root is$\: x=\frac{a}{b}\:$with$\:gcd(a,b)=1\: ($meaning the greatest common divisor of$\: a \:$and$\:b \;$is$\: 1.)$

$$Substituting$\; x=\frac{a}{b}\; $into the equation you will see that$\; a\; $divides$\; a_{0}\; $and$\; b \;$divides$\; a_{n}.$

$$The special case of this theorem is you when$\; a_{n}=1.\; $(Integer root theorem). I can write a proof for you once I find a computer for now if you want you can think of it like this, any integer root of the polynomial above call it$\;x_{0}\;$divides&\; a_{0}$

$$So in the case above we look at factors of$\;2 \;$and substitute it into the equation to see if any are roots$.$

$So we can try:$\; x = -1, 1, -2, 2.\;$Turns out$\; x = -1\; $is a root.$ .

[HR][/HR]

$[Rational Root Theorem]$

$\mathit{Proof}:\; $Suppose we have some polynomial$\;P(x)= a_{n}x^n + a_{n-1}x^{n-1}+...+a_{0} \; $given that each rational root is$\: x=\frac{a}{b}. \;$We have$,$

$P(\frac{a}{b})= a_{n}(\frac{a}{b})^n + a_{n-1}(\frac{a}{b})^{n-1}+...+a_{0} = 0$

$a_{n}(\frac{a^n}{b^n}) + a_{n-1}(\frac{a^{n-1}}{b^{n-1}})+...+a_{0} = 0$

$$Multiplying both sides by$\; b^n,$

$a_{n}(a^n)+a_{n-1}a^{n-1}(b)+....+a_{0}b^n = 0$

$a(a_{n}(a^{n-1})+a_{n-1}a^{n-2}(b)+....a_{1}b^{n-1})=-a_{0}b^n$

$$We note on the LHS we have$\; a \;$times an integer. So we can say$\; a\; $divides$\; -a_{0}b^n. \;$It follows$\; a \;$divides$\; a_{0}.$

$$We argue in a similar way (by factoring b) that$\; b \;$divides$\; a_{n}.$

QZP · Oct 12, 2013

@Spiralflex
Thank you but I have not learnt enough polynomials to be able to understand. When I gain more knowledge I will look back

Verify · Oct 12, 2013

These proofs look scary :'(

RealiseNothing · Oct 12, 2013

Verify said:
These proofs look scary :'(

They are actually really easy. It's just that the notation is very intimidating. Get your head around it and you'll be sweet.

Complex numbers (1 Viewer)

yeyjasminetime

Member

Sy123

This too shall pass

yeyjasminetime

Member

Sy123

This too shall pass

SpiralFlex

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QZP

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SpiralFlex

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SpiralFlex

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QZP

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Verify

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RealiseNothing

what is that?It is Cowpea

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