# Factorising complex equations...

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Can anyone give any tips/techniques when factorising complext equations

e.g:

z^2+6z+13,

z^2+z+1,

z^3+1

Many Thanks

e.g:

z^2+6z+13,

z^2+z+1,

z^3+1

Many Thanks

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#2

Use the quadratic formula for the first 2, and for the 3rd, take out a factor of (z-1) first then it leaves another quadratic.

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#5

Here's a general method for solving equations like the 3rd one.

f(x) = x³+ax²+bx+c

Take the factors of c and use them in f(x). If the result is zero, that number is a factor.

eg:

f(x) = x³+1

Factors of one: 1 (and -1)

f(1) = 1+1 = 2, so this isn't a factor.

f(-1) = -1+1 = 0, so this is a factor.

So.. x=-1 => (x+1)=0

Now divide x³+1 by x+1 to get a quadratic equation, and factorize that to get all the required factors.

f(x) = x³+ax²+bx+c

Take the factors of c and use them in f(x). If the result is zero, that number is a factor.

eg:

f(x) = x³+1

Factors of one: 1 (and -1)

f(1) = 1+1 = 2, so this isn't a factor.

f(-1) = -1+1 = 0, so this is a factor.

So.. x=-1 => (x+1)=0

Now divide x³+1 by x+1 to get a quadratic equation, and factorize that to get all the required factors.

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#6

Since you're dealing with complex numbers, remember you can use the following identity,

z² + a² = (z + ia)(z - ia)

z² + a² = (z + ia)(z - ia)

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#7

(Original post by

Since you're dealing with complex numbers, remember you can use the following identity,

z² + a² = (z + ia)(z - ia)

**Fermat**)Since you're dealing with complex numbers, remember you can use the following identity,

z² + a² = (z + ia)(z - ia)

**equations**.

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#8

(Original post by

Can anyone give any tips/techniques when factorising complext equations

e.g:

z^2+6z+13,

z^2+z+1,

z^3+1

Many Thanks

**sweetymango**)Can anyone give any tips/techniques when factorising complext equations

e.g:

z^2+6z+13,

z^2+z+1,

z^3+1

Many Thanks

z^2+6z+13

= (a+bj)^2 + 6(a + bj) + 13

= a^2 + 2abj - b^2 + 6a + 6bj + 13

= (a^2 - b^2 + 6a + 13) + (2ab + 6b)j

Is that how you mean?

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#10

(Original post by

Complex

**shift3**)Complex

**equations**.*will*involve complex numbers, and should you have the complex equation,

z² + a² = w²

then this can be factorised and rewritten as,

(z+ia)(z-ia) = w²

==========

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#11

I think the person means complex as in difficult and not the mathematical sense Fermat...

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#12

(Original post by

I think the person means complex as in difficult and not the mathematical sense Fermat...

**ZJuwelH**)I think the person means complex as in difficult and not the mathematical sense Fermat...

I think I'll go back to sleep.

Silly Zzzzzzzz....

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#13

I've only ever seen z notation in the P4 complex numbers chapter, so surely Mr. Fermat was correct in his efforts?

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#15

Can anyone give any tips/techniques when factorising complext equations

e.g:

z^2+6z+13,

z^2+z+1,

z^3+1

e.g:

z^2+6z+13,

z^2+z+1,

z^3+1

**not**equations. Accepting the standard notation that z = x+iy where x,y are real numbers and i = sqrt(-1) then they can, correctly, be described as complex functions.

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to clear things up, i am using z as a complex number i.e in the form of x+yj

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(Original post by

z = a + bj

z^2+6z+13

= (a+bj)^2 + 6(a + bj) + 13

= a^2 + 2abj - b^2 + 6a + 6bj + 13

= (a^2 - b^2 + 6a + 13) + (2ab + 6b)j

Is that how you mean?

**benm**)z = a + bj

z^2+6z+13

= (a+bj)^2 + 6(a + bj) + 13

= a^2 + 2abj - b^2 + 6a + 6bj + 13

= (a^2 - b^2 + 6a + 13) + (2ab + 6b)j

Is that how you mean?

**is**to use the quadratic formula

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