how do you simplify (4+3i)(5-7i)?

Answer 1

(4 + 3i)(5 - 7i)
20 - 28i + 15i - 21i^2
20 - 13i - 21(-1)
20 - 13i + 21
41 - 13i

Answer 2

The FOIL method. First, Outside, Inside, Last. Just like any other polynomial simplification. Of course i = √(-1).

So it will look like this:
First Outside Inside Last
(4)(5) + (4)(-7i) + (3i)(5) + (3i)(-7i) =
20 - 28i +15i - 21i² =
20 - 13i + 21 =
41 - 13i (simplified)

Answer 3

Multiply it out using the distributive law, or with FOIL, remembering that by definition i^2 = -1.

(4+3i)(5-7i) = 4(5 - 7i) + 3i(5-7i) = 4*5 - 4*7i + 3i*5 - 3i*7i
=20 - 28i + 15i - 21i^2 = 20 -28i +15i - (-21) = 20 -28i +15i + 21

Now combine like terms:
20 -28i +15i + 21 = (20 + 21) + (15 - 28)i = 41 - 13i

Answer 4

a million. First multiply them mutually a similar way you may want to multiply any 2 binomials, multiply 4 and a couple of, and four and -7i, then multiply -8i and a couple of, and -8i and -7i 8 - 28i - 16i + 56i^2 combine - 28i and -16i, and keep in thoughts i = sq. rt. of -a million, so i^2 = -a million 8 - 44i - fifty six combine 8 and -fifty six -40 8 - 44i 2. 27^-(2/3) The adverse exponent merely places the large form 27^(2/3) (no extra adverse exponent) because the denominator of a unit fraction (a million because the numerator) so a million/(27^(2/3)) split both/3 into (27^(a million/3))^2 27^(a million/3) = dice root of 27 = 3 so 3^2 = 9 so the full component practice is a million/9 3. i^31 considering i = squarert.(-a million) i^2 = -a million going extra, in case you sq. that lower back, so i^4 (or (-a million)^2) = a million i^31 = i^28 * i^3 (considering 28 is the optimal numerous of four, the first 28 i's that you multiply mutually finally end up imparting you with a million. so, a million * i^3 = i^3 yet i^2 = -a million so i^3 = -1i -1i is the simplified answer. 4. only a subtraction/distribution challenge, you should distribute the minus signal, -8 - 5 and 8i - (-4i) -13 + 12i 5. (x^2/3)^(4/5) and exponent of and exponent, potential you may merely multiply the exponents mutually, 2/3 * 4/5 = 8/15 so, x^(8/15)

Answer 5

(4+3i)(5-7i)
= 4*5 +4(-7i) + 3i(5) +3i(-7i)
= 20 -28i +15i -21i^2
= 20 -13i +21 (since i^2 = -1)
=41-13i

Answer 6

(4 + 3i)(5-7i) =
20 - 28i +15i -21isquared =
20 -13i + 21isquared

(Sorry, I couldn't figure out how to superscript the "squared" part of the answer on the keyboard.)

Answer 7

FOIL - first-outside-inside-last
(4*5)+(4*-7i)+(3i*5)+(3i*-7i)
20+(-28i)+15i+(-21i^2)
20-28i+15i+21 ----> i^2=-1
20+21+15i-28i
41-13i

Answer 8

(4+3i)(5-7i)
=20-28i+15i-21i^2
=20-13i-21i^2
=20-13i-21(-1) since the imaginary number i^2=-1
=41-13i

Answer 9

(4+3i)(5-7i)
20 +15i - 28i - 21i ^ 2
20 - 13i - 21i ^ 2
20 - 13i - (21 * -1)
20 - 13i + 21
41 - 13i

Answer 10

i think it is hard to explain... multiply the 4 and 5 then the 3i and 5 and to the same to the 7i. it is hard to explain. hope that helped.