If are the solutions of :
By over C I mean it has a solution that is a complex number
................=(2z-2)^2 - (Squareroot of (13)*i)^2
................=((2z+2)+(Squareroot of 13 *i))((2z+2)-(Squareroot of 13 *i))
However, the real answer is (2z-1+4i)(2z-1-4i)
I can factorise questions such as z^2-4z+17 but when there is a number infront of z^2 I do not know how to do it.
Thank you for any help .
But one more thing. How did you chose the numbers 1 and 17 exactly. I know they work but how did you pick these 2.
I tried your method with 9z-24z+32
and came up with 9z-24z+16 +16
But it took me a while to come up with 16 and 16.
Is there a formula like the (x*0.5)^2 for the case where the coefficient of z^2 is 1, to work out 1 and 17 or 16 and 16.
I will lead you through both of those.
will factor with
I must end up with as my centre term after multiplying out the factors,
so that rules out the first pair if I'm looking for integers.
must give me
which is so now I see what my must be. I need there.
Then I just break up 17 into 16+1, since .
Another way to look at would have been to figure that 16 is the nearest square to 17
and so is a good place to start.
The most obvious place to start is
Hence we take 16 from 32 to get