Verified answer

Simplify (2 + 3i)(3 – 4i)?

You just simply multiply in:

= (6-8i+9i-12i^2) and i^2 = -1

= (6-8i+9i-12(-1))

= (6+12-8i+9i)

= (18+i)

Hope it helps.

W.

First, memorize the chant "first-inside-outside-last" - this is the order that you use for the multiplications.

using the chant the steps would be:

A) "first" - multiply the first number of each set of brackets

*in this case 2 x 3 = 6

B) "inside" - multiply the two numbers on the 'inside' of the equation (the two that would be next to each other if you didn't have the brackets around them)

*in this case 3i x 3 = 9i

C) "outside" - multiply the two numbers that are on the 'outside' of the equation (the two that are farthest apart)

*in this case 2 x -4i = -8i

(notice the negative sign stays with the 4 - if the equation were (3 + 4i) it would've been 2 x 4i = 8i, with the 'plus' implied - similarly, if it were (2 - 3i) then the 'inside' step would've been -3i x 3 = -9i)

D) "last" - multiply the last number in each set of brackets

*in this case 3i x -4i = -12i^2 (^2 is 'i 'to the power of 2, or ' i squared' - I don't know how to do that font in this format)

notice again the minus sign stays with the 4, and is seen in the result.

E) put it all together

(2 x 3) + (3i x 3) + (2 x -4i) + (3i x -4i)

= 6 + 9i + (-8i) + (-12i^2)

= 6 + 9i - 8i - 12i^2

= 6 + 1i - 12i^2

In case your teacher isn't great at explaining the basics, a couple things to keep in mind (a math genius could probably tell you why these rules work, but I'm not one so just memorized them like most everyone else):

i) if you multiply a negative number with a positive number, you get a negative result

ii) if you multiply two negative numbers or two positive numbers, you get a positive result

**teachers love putting in different signs to catch students not paying attention and 'dropping the negative'**

iii) when you multiple numbers with exponents (the 'power to' numbers) you add the exponents

e.g. i^2 x i^5 = i^7

using real numbers in an example 2^2 x 2^2 = 2^4 = 16 (you can work it out manually just to reassure yourself it works).

For multiplications like 2 x -4i it still works because there is really an unwritten 'i^0' with the two and an assumed '^1' on the i with the -4. (we don't write these in because it would just make things messy and not add any useful information - any number to the power of 0 = 1, so it'd be 2x1=2, and the ^1 on the -4 would also just add clutter).

I hope that made some sense.

Good luck!

multiply each term of the first bracket by each of the second as multiplication is distributative over addition.

= (2 x 3) + (2 x -4i) + (3i x 3) + (3i x -4i)

= 6 - 8i + 9i -12i^2

= -12i^2 + i + 6

if i is a variable levae it at this point if this is advanced maths and i is the imaginary number √-1 then:

i^2 = -1 so:

= 12 + i + 6 = 18 + i

Since you are learning about imaginary numbers (i), I imagine that you know how to multiply 2 binomials. Just think of "i" as any other variable.

(2*3)+(2*-4i)+(3i*3)+(3i*-4i). (now just multiply the terms)

6-8i+9i-12i^2 (i^2= -1)

6-8i+9i-12(-1) (multiply the 12 and the -1)

6-8i+9i+12

18-8i+9i (combine like terms)

18+i (no way to get rid of the i, this is your answer)

Hope this helps

First, you distribute the 2 that is outside the parentheses on the second value. You then have: 5 - 4i + 2 + 4i The (- 4i) and (+ 4i) cancel, so you have 5 + 2. So your answer is: 7.

Multiply all the way across starting from the first #.

2x3 = 6

2x(-4i) = -8i

Now move to the next # in the first parentheses:

3ix3 = 9i

3ix(-4i) = -12i²

Now add all the answers together:

6-8i+9i-12i²

Finish off the addition;

6+i-12i²

(2 + 3i)(3 – 4i) = 6 - 8i + 9i - 12i² = 6 + i - 12(- 1) = 18 + i

Alejandra

(2+3i)(3-4i)

+4i +4i [to cancel it out]

(2+7i)(3)

-2 -2

7i(1)

7i

i think