This seems so easy, but I completely forgot how to do it.
The problem is: (3 - 2i)(4 + i)
How would you solve this problem?
The problem is: (3 - 2i)(4 + i)
How would you solve this problem?
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FOIL: First Outer Inner Last
3*4+3*i+(-2i)*4+(-2i)*i
=12+3i-8i+(-2)*(-1)
=12-5i+2
=14-5i
3*4+3*i+(-2i)*4+(-2i)*i
=12+3i-8i+(-2)*(-1)
=12-5i+2
=14-5i
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Use FOIL it's an acronym. It stands for: F.=first so you multiply 3*4 which equals 12. O=outside. So you multiply 3*i which equals 3i. I= inside. So you multiply -2i*4 which equals -8i. Finally L=last so you multiply -2i*i which equals -2i^2 (^2 means squared.)
So you have -2i^2 -8i +3i + 12 so combine like terms. The only like terms you have are -8i and 3i because they both have 'i' at the end. So your final answer is: -2i^2 -5i +12. Hope this helps! :)
So you have -2i^2 -8i +3i + 12 so combine like terms. The only like terms you have are -8i and 3i because they both have 'i' at the end. So your final answer is: -2i^2 -5i +12. Hope this helps! :)
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ok... first you distribute. 3 times 4 and 3 times i which gives you 12 + 3i. next you distribute -2i times 4 and -2i times i which is gonna give you -8i -2i sqaured. so you get 12+3i-8i-2i sqaured. combine like terms and you get 12-5i-2i sqaured. Now since i squared = -1, -2i sqaured is going to just be 2. so then you have 12-5i+2. combine like terms again (12+2) and you get 14-5i
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use FOIL. (3*4) + (3*i)-(2i+4)-(2i*i)....12+3i-8i-2i(squar… Answer: 12+11i-21(squared)
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3 * 4 + 3 * i -8i + 2i^2
= 12 +3i - 8i -2
= 10 -5i
= 12 +3i - 8i -2
= 10 -5i
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Make a 2x2 box!
Put the numbers on the outside and multiply.
Put the numbers on the outside and multiply.
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If you really need math help check out con academy.com ( I forgot how to do that too haha )