Question

Help factor!!! (x-1+i√3)(x-1-i√3)

Answer 1

This is already factored. Do you need to multiply together?

Answer 2

Yes

Answer 3

I know how to use the foil method with 2 terms, but not with 3 like this

Answer 4

$$
(x-1+\sqrt{3}i)(x-1-\sqrt{3}i)\\
\text{keep in mind that => }(a-b)(a+b) = (a^2-b^2)\\
(\color{red}{x-1}+\color{blue}{\sqrt{3}i})(\color{red}{x-1}-\color{blue}{\sqrt{3}i})
$$

Answer 5

I need it all multiplied together.

Answer 6

Not like that lol

Answer 7

with this, it is easiest to think of the real part and the imaginary parts as "a" and "b"
in (a+b)(a-b)

Answer 8

You can try multiplying every term of the left factor by every term of the right factor. Then collect like terms and simplify.

An easier way will be if you notice that this is the product of a sum and a difference by rewriting it as:

\( [(x -1) + i \sqrt{3}) ] ~[(x - 1) - i\sqrt{3} ) ] \)

Answer 9

Ok one second

Answer 10

The product of a sum and a difference.

\( (a + b)(a - b) = a^2 - b^2 \)

\( [(x -1) + i \sqrt{3}) ] ~[(x - 1) - i\sqrt{3} ) ]\)

\(= [(x -1)^2 - (i \sqrt{3})^2 ] \)

Answer 11

I got \[x ^{2}-x-ix \sqrt{3}-x+1+i \sqrt{3}+ix \sqrt{3}-i \sqrt{3}-i ^{2}\sqrt{9}\] all multiplied out. Now I need to simplify I guess

Answer 12

yes, remember i^2 = -1

Answer 13

Good so far.

Answer 14

but if you look at mathstudent's post, you could do it this way

(x-1)^2 - i^2 (sqrt(3))^2
the i^2 is -1 and sqrt(3) squared is 3 so

(x-1)^2 + 3

which can now be expanded (relatively) easily
x^2 -2x+1 + 3
x^2 -2x +4

Answer 15

Well, I'm down to \[x ^{2}-2x+2\sqrt{9}\]

Answer 16

The square root of 9 is 3, so would that be \[x ^{2}-2x+2*3\] ?

Answer 17

To get the final answer of \[x ^{2}-2x+6\]?

Answer 18

you should have gotten down to
\[x^2 -2x + 1 + \sqrt{9} \]

double check what you did

Answer 19

1-\[i ^{2}\]= 1-(-1) = 1+1 = 2

Answer 20

I'm pretty sure it is 2

Answer 21

yes but it is 1 - i^2 * sqrt(9)

which is 1 - -1*sqrt(9) = 1 + sqrt(9)

Answer 22

It's a double negative so it turns to a positive

Answer 23

\((x-1+i\sqrt{3})(x-1-i\sqrt{3}) \)

\(x ^{2}-x-ix \sqrt{3}-x+1+i \sqrt{3}+ix \sqrt{3}-i \sqrt{3}-i ^{2}\sqrt{9}\)

\(x^2 - x - x -ix \sqrt{3} + ix \sqrt{3} + 1 + 3\)

Answer 24

yes but it is not 1 - i^2

it is
\[ 1 - i^2 \sqrt{9} \]

Answer 25

Which is \[1-(-1)\sqrt{9}\] lol

Answer 26

order of operations means you do i^2 to get -1
then multiply -1* sqrt(9) to get -sqrt(9)

now you have
1 - -sqrt(9)
which becomes
1 + sqrt(9)

Answer 27

Ohhhhhh ok ok. Wasn't thinking in terms of orders of operations for some reason haha

Answer 28

So the final answer would be \[x ^{2}-2x+4\]

Answer 29

yes. but if you go back to the previous posts you will see the fastest way to do this
is treat the real part (x+1) and the imaginary parts as (a+b i) (a- bi)
these are *complex conjugates* and you get a^2 - i^2 b^2 which is
a^2 + b^2

try to memorize that: complex conjugates give you the sum of the squares (pure real)

so
(x+1)^2 + 3
which becomes
x^2 -2x +4

Answer 30

Yeah, that would have saved me a lot of time. Thanks so much

Answer 31

it is worth doing it the hard way... it motivates learning the short cut.

Answer 32

So true! Have a great rest of the day