Factoring with complex numbers

[jpanknin]

Can someone explain why, when factored, both terms are ```x -``` (circled in red). I got the x term just fine, but am confused as to why both are minus signs after the x.

 

[jpanknin]

And same thing with this one:



This would be ```[x - (-1 + i)] [x - (-1 - i)]```. So again, why the minus after the ```x``` rather than a plus or a plus/minus?

 

[Zermelo]

Multiply the factored expression, and you'll see that it works out. That's the same way you would factor a quadratic equation with real solutions! x^2 -5x + 6 ---> solutions are 3 and 2, so the factored expression is (x -3)(x-2).
The intuition behind this is that solutions should be the roots (zeros) of the function. So, if you plug in your solution to the function, it should output 0. Similarly, if it were (x + sol1)(x + sol2), where sol1 and sol2 are the solutions, when you plug in sol1 for example you would get 2* sol1 (sol1 + sol2), which is not zero! That's why it is (x - sol1)(x - sol2), now if you put in x = sol1, you would get (sol1 - sol1)(sol 1 - sol2) which is zero. Hope I understood your question and gave a satisfying answer, be free to ask for more details if you are still not sure. Cheers

 

[Davey Jones]

Zermelo summarized it nicely. Your solution should take a factor and make it equal zero. The easiest way to make zero is to subtract something from itself. For example, if you have a solution to a polynomial equation of x=3, then substituting this into (x-3) with make the factor equal to zero (and being a factor, aka part a multiplication problem), it makes the entire expression equal to zero.

And, it doesn't matter how ugly the solution is:
x=3? Then use (x-3) as a factor, since 3-3=0
x=SQRT(7)? Then use (x - SQRT(7)) as a factor, since SQRT(7) - SQRT(7) = 0
x = 3-4i? Then use (x-(3+4i)) as a factor, since (3+4i)-(3+4i) = 0.

Hope this helps

 

[Steven G]

If x=a, then x-a must be a factor!
Let the symbol q_i represent any number other than a. So if f(x) factors into f(x) = (x-q₁)(x-q₂)...(x-q_n), then f(a) can't be 0, since each factor, (x-q_i) becomes (a-q_i) which is not 0. Therefore, for a to be a root/zero/solution, then (x-a) must be a factor.

 

[JeffM]

Can someone explain why, when factored, both terms are x−x- (circled in red). I got the x term just fine, but am confused as to why both are minus signs after the x.

View attachment 26801

It is a theorem about polynomials that x minus any root of f(x) is ALWAYS a factor of f(x) if f(x) is a polynomial.

[MATH]f(x) =ax^2 + bx + c, \ u = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a}, \text { and } v = \dfrac{-b - \sqrt{b^2 - 4ac}}{2a}.[/MATH]
[MATH]\therefore f(u) = a \left ( \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \right )^2 + b \left ( \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \right ) + c =[/MATH]
[MATH]a \left ( \dfrac{b^2 - 2b\sqrt{b^2 - 4ac} + b^2 - 4ac}{4a^2} \right ) + \dfrac{-b^2 + b\sqrt{b^2 - 4ac}}{2a} + c =[/MATH]
[MATH]\dfrac{2b^2 - 2b\sqrt{b^2 - 4ac} - 4ac}{4a} + \dfrac{-2b^2 + 2b\sqrt{b^2 - 4ac}}{4a} + \dfrac{4ac}{4a} =[/MATH]
[MATH]\dfrac{2b^2 - 2b^2 + 2b\sqrt{b^2 - 4ac} - 2b\sqrt{b^2 - 4ac} +4ac - 4ac}{4a} = \dfrac{0 + 0 + 0}{4} = 0.[/MATH]
I shall let you satisfy yourself that f(v) = 0.

[MATH]a \left ( x - \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \right ) \left ( x - \dfrac{-b - \sqrt{b^2 - 4ac}}{2a} \right ) =[/MATH]
[MATH]ax^2 - \dfrac{-bx + x\sqrt{b^2 - 4ac}}{2} - \dfrac{-bx - x\sqrt{b^2 - 4ac}}{2} + \dfrac{b^2 - b^2 + 4ac}{4a} =[/MATH]
[MATH]ax^2 - \dfrac{2bx}{2} + \dfrac{4ac}{4a} = ax^2 + bx + c.[/MATH]
It is a specific application of the Fundamental Theorem of Algebra to quadratics.

 

[HallsofIvy]

If x= a then x- a= 0. So (x- a) times anything will be 0 when x= a.

 

[Steven G]

If x= a then x- a= 0. So (x- a) times anything will be 0 when x= a.

Yes, but must (x-a) be a factor?! You never mentioned that.

 

[HallsofIvy]

Yes, but must (x-a) be a factor?! You never mentioned that.

A factor of what? The original post asked (I am using "a" and "b" instead of the numbers used) "If a and b are roots of of a quadratic function x- a and x- b are factors. Why is "-"? Why not "+"? That was the question I was addressing.

 

[Steven G]

A factor of what? The original post asked (I am using "a" and "b" instead of the numbers used) "If a and b are roots of of a quadratic function x- a and x- b are factors. Why is "-"? Why not "+"? That was the question I was addressing.

OK, Dr Halls!