help with an algebra question.

[the game]

So I'm working on my algebra homework and not sure why I have this problem wrong. Can somebody help me out with it? Preferably show steps. Below is what I have.

i(3-4i)(3+4i)
3i-4i+3+4i
3+3i

[Kujako]

Try working with Wolfram Alpha. Better option then just asking for answers.

[wregelmann]

You turned multiplication into addition, and made i*4i = 4i. The second line should read (3i+4)(3+4i). Try it from there.

[Vizardlorde]

wth did you do in that second step
if you are going to expand it should be i(9+12i-12i+16)
25i if im not mistaken

[Flabberly]

Dont know how you arrived at your conclusion. The correct one is:
i(3-4i)(3+4i) <- separate the binomial
i(9+12i-12i-16i^2) <-delete both 12i since they are complementary and i^2=-1
i(9+16) <- simple from here
25i

[unbound]

So I'm working on my algebra homework and not sure why I have this problem wrong. Can somebody help me out with it? Preferably show steps. Below is what I have.

i(3-4i)(3+4i)
3i-4i+3+4i
3+3i

i(9-12i+12i-16i^2)

i(9-16(-1))

i(9+16)

25i

[Garnier Fructis]

For any complex number a + bi, the following is true:

(a + bi)*(a - bi) = a^2 + b^2.

So you get i*(3^2 + 4^2). To be honest I'm not completely clear how you did your calculation. You seem to have commuted the i with the left parentheses and turned 3 into 3i, and then you turned multiplication into addition. Neither of which are valid manipulations.

Dont know how you arrived at your conclusion. The correct one is:
i(3-4i)/(3+4i) <- separate the binomial
i(9+12i-12i+16i^2) <-delete both 12i since they are complementary and i^2=-1
i(9-16) <- simple from here
-7i

i^2 is -1, so you made a mistake going from line 2 to 3. Also, you made a typo in the first line by writing division.

[stormgust]

Uhm, use the binomial theorem, which holds for all commutative rings (i.e. also the complex numbers). and i^2=-1. Should do the trick (25i is correct)

[lockedout]

i * (3 - (4 * i)) * (3 + (4 * i)) = 25 i

[the game]

i(9-12i+12i-16i^2)

i(9-16(-1))

i(9+16)

25i

So my whole problem was distributing the first I too early?

[dawawe]

i(3-4i)(3+4i)
(3i-4i^2)(3+4i)
9i+12i^2-12i^2-16i^3
9i-16^3

[MrKnubbles]

So I'm working on my algebra homework and not sure why I have this problem wrong. Can somebody help me out with it? Preferably show steps. Below is what I have.

i(3-4i)(3+4i)
3i-4i+3+4i
3+3i

i(3 - 4i)(3 + 4i) There are a couple ways to do this but here is one way.
i(9 + 12i - 12i - 16i^2) Ignore the i and FOIL the parentheses.
i(9 - 16i^2) Simplify to get this. Final answer.
9i - 16i^3 If they want the expression completely merged together then this is the final answer.

Are you working with imaginary numbers? Meaning that i = sqrt(-1) because if you are then that changes things a bit.

[the game]

Thanks guys, I understand where my mistake was now.

[Stop Pretending]

So my whole problem was distributing the first I too early?

Should have been distributing the D instead.

[MAGAmobile]

So my whole problem was distributing the first I too early?

The main problem with that step in general is that you didn't distribute i fully to begin with. Foiling out first would be easier.

[stormgust]

So my whole problem was distributing the first I too early?

You can distribute it in any step you want, if you do it correctly.

pulling the i into the first bracket it becomes:
(3i-4i^2)(3+4i)=(3i+4)(3+4i)=9i+12+12i^2+16i=25i

or by pulling the i into the 2nd bracket:
(3-4i)(3i-4)=(9i+12-12+16i)=25i.

For some reason you turned the 2nd mupliplication into an addition after distributing the i onto the first factor.

[Jonnusthegreat]

As stated, the second and third term make a conjugate. With conjugates (a+bi)(a-bi), the middle term cancels so it forms a real answer.

(3+4i)(3-4i) = 9 - 16i^2 = 9 - -16 = 9 + 16 = 25

Then multiply by the first term and you get 25i.

[Garnier Fructis]

Uhm, use the binomial theorem, which holds for all commutative rings (i.e. also the complex numbers). and i^2=-1. Should do the trick (25i is correct)

I'm pretty sure Ring theory is really far from whatever he's doing, lol.

[Vizardlorde]

So my whole problem was distributing the first I too early?

No you didnt distribute it properly.

[Planeshaper]

So my whole problem was distributing the first I too early?

You can distribute in whichever order you want, but you didn't distribute correctly:

i(3-4i)(3+4i) ==

(3i+4)(3+4i) - i distributed into first parenthetical.

(3-4i)(3i-4) - i distributed into second parenthetical.

i(9-12i+12i+16) - first and second parentheticals distributed into each other.

All yield the answer 25i, which is the answer.