Simpify Factored Form Equation

[kato1]

Simplify the function f(x)= (x-(1+i))(x-(1-i)) .

Is the function you wrote in factored form [ f(x)= (x-(1+i))(x-(1-i)) ] , the same as the original function in
standard form [ f(x) = x^2 - 2x +2) ] ?

 

[MarkFL]

Hello and welcome to MHB! (Wave)

We typically like for our users to post their work so we can see where you are stuck and what you've tried. That being said, let's examine the function:

\(\displaystyle f(x)=x^2-2x+2\)

By the quadratic formula, we obtain the roots:

\(\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(2)}}{2(1)}=\frac{2\pm\sqrt{-4}}{2}=1\pm i\)

Therefore, we know:

\(\displaystyle f(x)=x^2-2x+2=(x-(1+i))(x-(1-i))\)

Now, if we simply wish to verify this, we may take the factored form and expand it as follows:

\(\displaystyle (x-(1+i))(x-(1-i))=x^2-x(1-i)-x(1+i)+(1+i)(1-i)=\)

\(\displaystyle x^2-x+ix-x-ix+1-i^2=x^2-2x+2\quad\checkmark\)

 

[kato1]

Hello and welcome to MHB! (Wave)

We typically like for our users to post their work so we can see where you are stuck and what you've tried. That being said, let's examine the function:

\(\displaystyle f(x)=x^2-2x+2\)

By the quadratic formula, we obtain the roots:

\(\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(2)}}{2(1)}=\frac{2\pm\sqrt{-4}}{2}=1\pm i\)

Therefore, we know:

\(\displaystyle f(x)=x^2-2x+2=(x-(1+i))(x-(1-i))\)

Now, if we simply wish to verify this, we may take the factored form and expand it as follows:

\(\displaystyle (x-(1+i))(x-(1-i))=x^2-x(1-i)-x(1+i)+(1+i)(1-i)=\)

\(\displaystyle x^2-x+ix-x-ix+1-i^2=x^2-2x+2\quad\checkmark\)

Hello. I'm confused as to how you get from x^2-x+ix-x-ix+1-i^2 to x^2-2x+2 . Can you please elaborate?
So far I've tried simplfying up to:
=x^2-x+ix-x-ix+1-i^2
= x^2-(x-x)+(ix-ix)+(1- √ ̅-1 )
= ?

Note: I substituted i^2 for √ ̅-1 .
Is (1- √ ̅-1 ) =2 ? And how? Thank you again.

 

[MarkFL]

Note that:

\(\displaystyle i\equiv\sqrt{-1}\therefore i^2=-1\)

And so:

\(\displaystyle x^2-x+ix-x-ix+1-i^2=x^2-x(1+1)+ix(1-1)+1-(-1)=x^2=x^2-2x+2\)