Simplifying help needed?

Question

I need it in the form a + bi

I mistyped this earlier...... the i's are i as in -1, not the letter i

a) (2-3i)^2
b) (2i+3)(3i^3 - 4i^2 + 7)
c) (2+3i)/(1-2i) rationalize

I'm not just looking for the answers. I'm hoping someone can provide those, but also the work used to get them, so I can visualize what you did.

Thank you so much! I appreciate your time.

Please assist.

Answer 1

(2-3i) ^2
4-12i-9i^2
4-12i+9
13-12i

(2i+3)(3i^3-4i^2+7) since 3i^3 = -3i -4i^2 = 4
(2i+3)(-3i+4+7)
(2i+3)(-3i+11)
-6i^2 +22i -9i +33
6 +22i +33
22i+39

(2+3i)/(1-2i)
(2+3i)/(1-2i) * (1+2i)/(1+2i)
(2+4i+3i+6i^2)/(1-4i^2)
(2+7i-6)/(1+4)
(-4+7i)/5

Answer 2

just remember that i^2 = -1

1. (2 - 3i)(2 - 3i) = 4 - 12i + 9i^2
since i^2 = -1:
4 - 12i + 9i^2 = 4 - 12i + 9(-1) = -12i -5 = - (12i + 5)

2. (2i+3)(3i^3 - 4i^2 + 7) = 6i^4 - 8i^3 + 14i + 9i^3 - 12i^2 + 21
since i^2 = -1:
6i^4 - 8i^3 + 14i + 9i^3 - 12i^2 + 21
= 6(-1)(-1) - 8(-1)i + 14i + 9(-1)i - 12(-1) + 21
= 6 + 8i + 14i - 9i + 12 + 21
= 39 + 13i
= 13 (3 + i)

3. this requires rationalising as you've already mentioned. the point of rationalising is to remove the i from the denominator and the only way to do that is to form a (a + b)(a - b) equation:
(2+3i)/(1-2i) multiply (1+2i)/(1+2i)
= [(2+3i)(1+2i)] / [(1-2i)(1+2i)]

do you see that the denominator has formed an (a + b)(a - b) equation? this will cancel out the i and you will not be left with any at the denominator, only at the numerator

after looking through questions 1 and 2 i believe that you can carry on with question 3 by yourself. just remeber the golden rule that i^2 = -1 and you cannot go wrong.

good luck and have fun with your math!

Answer 3

A.) First get rid of the -1. Multiply 3 by -1. You will get positive three. then do 2+3. that will give you 5. then do 5 to the square root of 2. that will give you 25. that is your answer for A. I think that is what you are trying to do.

B.) First get rid of the -1. So you do 2*-1. Which will give you -2. Then you get rid of all the other -1s in the other equation. So you would do 3*-1^3-4*-1^+7. So 3*-1 is -3. -4*-1 is 4. So now your equation is (-2)(-3^3+4^2+7) so now get rid of the square roots. So -3^3 is -27. Then 4^2 is 16. so now you have (-2)(-27+16+7). Now you add them up. You so 16+7=23. Then do -27+23=-4. Now you do -2*-4=8. So your answer is 8.

C.) For this one i think that you just multiply. You can do 2*1 which will give you 2. Then do -3*2=6 so Now you have (2+6) which will also give you 8. That is how i think you do C.

hope i helped out.
Good Luck!!!

Answer 4

I'll start by saying that I'm a little rusty. And I apologize if I end up sounding like I'm condescending. The easiest way I know to explain things is to break it all the way back to the basic steps. I also don't know you, so I don't know what kind of background you have, or which parts are easiest to understand and which parts hardest to understand (where you excel and where you struggle). Please don't be offended.
a) (2-3i)^2
Rewrite it to look like (2-3i)(2-3i), then proceed to FOIL it (First Outside Inside Last). (my teachers drilled that into my head so well, that I can't look at these kinds of problems without thinking the word "foil")
Now we multiply the terms in FOIL order: start by multiplying 2*2 (the First terms) [4], then 2*(-3i) (the Outside terms) [-6i], then (-3i)*2 (the Inside terms) [-6i], then (-3i)*(-3i) (the Last terms) [-9] (that answer comes from -3*-3=9, and i*i=-1, then 9*-1=-9)
Those answers end up looking like this: 4-6i-6i-9. And we combine any whole numbers together, and any like-term items together: 4-9 [-5] and -6i-6i [-12i], and rewrite it in "a+bi" format: -5-12i.
b) (2i+3)(3i^3-4i^2+7)
We'll start by simplifyng the Imaginary terms and rewriting the equation. This could be solved first by doing the FOIL thing on a larger scale, and then simplifying the Imaginary terms, but this let's us start with simpler terms, and doesn't get as tricky. I'll show this way without the explanations afterward. Remember that i^3=-i, and i^2=-1: (2i+3)(-3i+4+7) or (2i+3)(-3i+11)
First terms: (2i)*(-3i)=6 (2*-3=-6, i*i=-1, -1*-6=6)
Outside terms: (2i)*(11)=22i
Inside terms: 3*(-3i)=-9i
Last terms: 3*11=33
Rewrite these answers together: 6+22i-9i+33. Combine like terms: 39+13i (33+6=39 & 22i-9i=13i) and it's done. Now, I'll just write out the other method where I FOIL first, and simplify after, without the explanations:
(2i+3)(3i^3-4i^2+7) =
6i^4-8i^3+14i+9i^3-12i^2+21 =
6+8i+14i-9i+12+21 =
6+12+21+8i+14i-9i =
39-13i
c) rationalize (2+3i)/(1-2i)
Since any fraction can be multiplied by 1, and still be equal to itself, we multiply by a "well-chosen 1", as my teacher used to say: In this case, we want to end up with something as small on the bottom as possible (like what Hollywood says about women).
Going back to some of the Polynomials that are easily factored, we can remember the Difference of Squares: x^2-y^2, which factors out as (x+y)(x-y). Looking at the similarities, we can see that the bottom of this fraction (1-2i) looks a bit like the (x-y) factor in the DIfference of Squares example. What we're missing is the (x+y) factor, or, in our case, (1+2i). Since (1+2i)/(1+2i)=1 (anything divided by itself equals 1), we can multiply our fraction by (1+2i)/(1+2i); or to rewrite it, (2+3i)*(1+2i)/(1-2i)*(1+2i).
Now, we can start FOILing to get to some terms we can reduce.
The bottom (taking our cue from the Difference of Squares, and working backwards), we know (1-2i)*(1+2i) = 1*1+1*2i-2i*1-4i^2,
since i^2=-1, we can also write this as 1+2i-2i+4, or 5 (when all the arithmetic is followed through, 1+4, and 2i-2i)
Now, we have the top: (2+3i)*(1+2i)
First terms: 2*1 = 2
Outside terms: 2*2i = 4i
Inside terms: 3i*1 = 3i
Last terms: 2i+3i = 6*i^2 = 6*-1 = -6
These are rewritten together as 2+4i+3i-6, or 2-6+4i+3i, which simplifies down to -4+7i.
So, when we put these pieces back together, we have:
(-4+7i)/5. that's as rational as this problem gets.

I hope this helps, and good luck. Let me know if I was any use to you, at all.

Answer 5

just treat the i as a letter, do your calculations and replace the i*i terms with -1. that is all.

for c) multiply with the (1+2i)/(1+2i) the denominator will become a real number instead of a complex one.

Answer 6

Simplify what's interior the brackets first. (Multiply 2 * x and a pair of * a million) = (2x+2-7)=(2x-5) Now place parenthesis around the two words and FOIL (6x-2)(2x-5)= (12x^2 - 30x-4x+10)=(12x^2-34x+10) ^2 means squared FOIL means First, exterior, interior, final wish this allows!

Answer 7

kish is correct bob h is wrong atleast in answer 1.

Answer 8

Please do your homework yourself.

Why not ask a fellow student or your teach for help if you're stuck.