how do you multiply these two equations?

Question

how do you multiply these two equations...what is the strategy?
[x-(1+3i)] [x-(1-3i)] Appreciate it!

Answer 1

The same way as any two binomials

x² + (x - 2xi) + (x + 2xi) + [1 - 3i + 3i -9i²]
x² + 2x + 1 - 9
x² + 2x - 8

Answer 2

The strategy is this: multiply first terms, then note the alternate terms are 2(1 + 3i)(1 - 3i) and the last terms become the old "difference of squares". So you have x^2 + 2(1 + 3i)(1 - 3i) + (1^2 - 9i^2) [the middle terms cancel out]. Now the one thing to recall at this point is that i^2 = -1, so instead of 1 - 9i^2, we have 1 - 9(-1) = 1 + 9 = 10. The final equation looks like this: x^2 + 2(1 +3i)( 1 - 3i) + 10. The fact that (1 + 3i)(1 - 3i) appears in the middle is the indication that you can factor this equation and that 10 breaks down to these conjugates in the middle terms. I leave you to the simplification of the last equation.

Answer 3

you multiply

x^2 -x (1+3i) -x(1-3i) +(1+3i) (1-3i)

remark that (1+3i) (1-3i) = 1-9i^2 and since i^2 =-1 =10

x^2 -x (1+3i+1-3i) +10

x^2-2x-10

Answer 4

x/29 + ([4/5][y/29]) = a million x/29 + 4y/one hundred forty 5 = a million 5x + 4y = one hundred forty 5 4y = one hundred forty 5 - 5x y = (one hundred forty 5 - 5x)/4 x/32 + ([9/10][y/32]) = a million x/32 + 9y/320 = a million 10x + 9y = 320 9y = 320 - 10x y = (320 - 10x)/9 locate x: 9(one hundred forty 5 - 5x) = 4(320 - 10x) a million,305 - 45x = a million,280 - 40x 5x = 25 x = 5 locate y in 1st equation: = (one hundred forty 5 - 5[5])/4 = (one hundred forty 5 - 25)/4 = 100 and twenty/4 or 30 answer: x = 5, y = 30 info (second equation): 5/32 + ([9/10][30/32]) = a million 5/32 + 27/32 = a million 32/32 = a million a million = a million

Answer 5

Multiply each part of it
(a1 + b1) (a2 + b2) = a1a2 + a1b2 + a2b1 + b2b1

= x.x + x.(-(1-3i)) + x.(-(1+3i)) + (-(1-3i).-(1+3i))
= x2 + (-x + 3xi) + (-x – 3xi) + { (-1.-1) + (-1.-3i) + (-1.3i) + (3i.-3i)}
Feel free to continue

Answer 6

There is this formulae....
(a+b)(a-b) = a(squared) - b(squared)

so:
(x-1-3i) (x-1+3i)
=x(x)[-1(squared) -3i(squared)]
=x(squared) -1(squared) -3i(squared)


Hope I got it right ;)

Answer 7

These are the factors of a quadratic expression. The roots are complex conjugates and hence the expression will have real co-efficients,
Expn is x^2 - S + P,
S = sum of roots & P = product of roots.
x^2 - 2 + 4