(8i-3)(5i+2)?

Question

operations with complex numbers

Answer 1

(8i-3)(5i+2)
First use FOIL(First, Outer, Inner, Last)
8i x 5i(F)=40i^2
8i x 2(0)=16i
-3 x 5i(I)=-15i
-3 x 2(L)=-6
combine like terms
40i^2+i-6

i^2=-1

40(-1)=-40
combine like terms
i-46
put it into this from: a+bi
-46+i or -46+1i

Answer 2

Keep in mind that a complex number is of the form (a + bi), for some real constants a and b, and where i = sqrt(-1).

(8i - 3) (5i + 2) represents the multiplication of two complex numbers.

To multiply, all we have to do is multiply this as we would a binomial, using FOIL (first, outer, inner, last).

(8i)(5i) + 2(8i) - 3(5i) - 6. Multiplying the terms together, we get
40i^2 + 16i - 15i - 6. Grouping like terms,
40i^2 + i - 6

And now, remember that i^2 is equal to (-1). We replace the i^2 with (-1)

40(-1) + i - 6. From more simplification,
-40 + i - 6. And, from combining the constant terms,
-46 + i

Note that this is certainly of the from a + bi, with a = -46 and b = 1.

Answer 3

(8i-3)(5i+2) use FOIL

40i^2+16i-15i-6
-40+i-6
-46+i

Answer 4

Multiply the numerator and denominator by the conjugate of the denominator (4 + 5i) (7 - 8i)(4 + 5i)/(4 - 5i)(4 + 5i) = (28 - 32i + 35i - 40i^2)/(sixteen - 25i^2) = (28 + 3i + forty)/(sixteen + 25) (sixty 8 + 3i)/forty-one --> sixty 8/forty-one + 3/forty-one i

Answer 5

FOIL, and remember that i^2 = -1
-40 -15i + 16i - 6
-46 + i

Answer 6

Does this have anything to do with i?
Either way, you still have to use FOIL.
(8x5i)+(8ix2)=40i^2+16i
(-3x5i)+(-3x2)= -15i-6
40i^2+i-6 << Since i^2=-1, then (40)(-1)= -40
-40+i-6
-46+i<<

Answer 7

=-40+i-6
=-46+i

Answer 8

40i^2 -15i +16i -6

-40 +i -6
-46 +i

Answer 9

-40 +16i-15i-6

=-46+i

Answer 10

foil
40i^2+16i-15i-6
-40+i-6
-46+i