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x^2 + 64 this is a like the difference of 2 squares except you have to use the imaginary number i remember that i^2 = -1 so can you fit the i values in here x^2 + 64 = ( x + ____)( x - ____)
im not sure sorry
ok what is the square root of 64?
ok so expand this and see if it comes back to x^2 + 64 (x + 8i)(x - 8i)
just as in difference of 2 squares the middle terms will cancel out
okayy.... so that is the first one
(x + 8i)(x - 8i) = x^2 + 8ix - 8ix -64*-1 = x^2 + 64
16x^2 + 49 handle it in the same way
take square root of 16x^2 and 49
No - where are the i's???
and you take Square root of 16x^2
well its 4, but idk what to do with it
it 4x 4x * 4x = 16x^2
put 4x in each bracket
yes (16x + 7i)(x - 7i) would not come back to 4x^2 + 64 because the 2 terms in ix would not cancel out
okay i get it
next ones a bit different
Oh no thats not quite right Look at your signs - doddnt forget its like difference of 2 squares
NO its 4x !!!
(4x + 7i)(4x - 7i)
sorry lol i copy and pasted the wron egquation
i wanted to copy then change the signs but i copied the wrong one
dont know why but i can't copy and paste!
Find the product of (x + 9i)^2. next one
withe the last 3 you just expand the parentheses in the normal way (x + 9i)(x + 9i) first multiply the second bracket by x then multiply it by 9i = x(x + 9i) + 9i(x + 9i) = x^2 + 9xi + 9xi + ? can you do the last bit?
9i*9i = ?
sorry i really tired today
81* i^2 and what is i^2 equal to?
so that is 81 * -1 = -81 x^2 + 9xi + 9xi - 81 now just add the 2 middle like terms
right giving x^2 + 18xi - 81
is that it?
Yes the 4 th one is done in exactly the same way
can u just set it up (expand it) and ill take from their then you check?
(x - 2i)(x - 2i) = x(x - 2i) - 2i(x - 2i)
right so up until the last term -2i*-2i is not -2i^2 - * - = + and 2*2 = 4
so -2i * -2i = ?
treat i as if it weer an x or a y x^x = x^2 i* i = ?
* tahs x * x = x^2
right + 4 i^2 = ?
what do you mean by equals?
i^2 = -1 so 4i^2 = ?
important to remember i = sqrt(-1) so i^2 = -1
x^2-2xi-2xi-4 = x^2 - 4xi - 4
so is that the answer cause it says to find the product
yes the last one is more complicated because the last term in the brackets is (3x + 5i) so you could do the last part (3x + 5i)^2 first then plug it in later
you can do this in the same way as the last 2 (3 + 5i)(3 + 5i) = 3(3 + 5i) + 5i(3 + 5i) ok try and do that
yea good job and we can simplify it further to 30i - 16
is that the answer to the problem?
No we have to plug that in to the expansion (x + (3+5i)(x + (3+5i) = x(x + (3+5i) + (3 + 5i)(x + (3+5i) = x^2 +3x + 5xi + 3x + 9 + 15i + 30i -16
hold on - I'm getting tired to thats wrong
x(x + (3+5i) + (3 + 5i)(x + (3+5i) = x^2 + 3x + 5xi + 3x + 5xi + 30i - 16
thats it now simplify that was a tricky one
is that iT?
Thanks thats it for now
good I think I'll take a break lol