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Using euclidean's algorithm, we see that

\(\gcd(n^2, n-12)\\=\gcd(n^2-n(n-12), n-12)\\=\gcd(12n, n-12)\\=\gcd(12n-12(n-12), n-12)\\=\gcd(144,n-12)\)

Therefore, 144 must be divisible by n-12 in order for n^2 to be divisible by n-12.

\(144=2^43^2\), so it has (4+1)(2+1) = 15 factors (which are \(2^03^0, 2^13^0, 2^23^0, 2^33^0, 2^43^0, 2^03^1, 2^13^1, 2^23^1, 2^33^1, 2^43^1, 2^03^2, 2^13^2, 2^23^2, 2^33^2, 2^43^2\)), which means that if you add 12 to any of the numbers mentioned above, you will get a number n that satisfies the condition above.

Can you finish it?

since the final polynomial is a fourth-degree polynomial, Q(x) must be quadratic, and since it is monic, Q(x) is also monic. Therefore, Q(x) can be expressed as x^2+cx+d, where c and d are real numbers. If you were to square it, you would get x^4 + 2 c x^3 + c^2 x^2 + 2 d x^2 + 2 c d x + d^2. Therefore, we can solve it using equations by matching the coefficients:

\(2c=1\\c=1\\2d+c^2=2d+1=-4\\d=-2.5\)

Now just match the coefficients once again to solve for a and b, and then add them up. Can you finish it?

Let z = a+bi, where a and b are real numbers. After substituting, we get:

\(|(a-3)+bi|=|a+(b+1)i|=|(a-1)+(b-2)i|\)

By the definition of the absolute value of a complex number,

\(\sqrt{(a-3)^2+b^2}=\sqrt{a^2+(b+1)^2}=\sqrt{(a-1)^2+(b-2)^2}\\ (a-3)^2+b^2=a^2+(b+1)^2=(a-1)^2+(b-2)^2\)

(we won't need to worry about extraneous solutions because everything will always be positive since it is the square of real numbers.)

\(a^2-6a+9+b^2=a^2+b^2+2b+1=a^2-2a+1+b^2-4b+4 \)

Subtract \(a^2+b^2\) from all equations:

\(\)\(-6a+9=2b+1=-2a+1-4b+4 \)

Solve it just like a normal system of equations:

\(-6a+9=2b+1\Rightarrow3a+b=4\\ 2b+1=-2a+1-4b+4\Rightarrow 3a+9b=6\\ 8b=2\\ b=\frac{1}{4}\\ 3a+\frac{1}{4}=4\\ a=\frac{5}{4} \)

Therefore, the solution to this system of equations is \(\boxed{\frac{5}{4}+\frac{1}{4}i}\)

\(|z|<8\) represents a circle with a radius of 8 centered at (0, 0), and the \(-3i\) simply shifts that circle upwards by 3 units. Therefore, the area is just the area of a circle with a radius of 8 and should be easily solvable by using the formula \(A=\pi r^2\), where A is the area of a circle and r is that circle's radius.

Can you finish it from there?

bruv why was this blocked lmao

anyway it's kinda useless cuz you can just open it on another window

The problem can be easily solved using the law of sines:

\(\frac{\sin(120)}{2\sqrt{3}}=\frac{\sin(45)}{x}\\ \frac{\frac{\sqrt{3}}{2}}{2\sqrt{3}}=\frac{\frac{\sqrt{2}}{2}}{x}\\ \frac{1}{2}=\frac{\sqrt{2}}{x}\)

Should be trivial by now. Can you finish it?

not even that hard to factor:

\(4x^4+12x^3-9x^2-27x\\=x(4x^3+12x^2-9x-27)\\=x(4x^2(x+3)-9(x+3))\\=x(4x^2-9)(x+3)\\=\boxed{x(2x+3)(2x-3)(x+3)}\)

btw if you didn't do it by hand then how did you do it?

if you were to just divide, you get 31743 / 281, which is approximately 112.96, which is very close to an integer. This suggests that if you increase 31743 by a few digits it will divide 281. Sure enough, when you increase it by 10, you get 31753, which is divisible by 281

I think the first part is meant to define the operation *, so it would be:

\(7+\sqrt{g+\sqrt{g+\sqrt{g+...}}} = 19\\ \sqrt{g+\sqrt{g+\sqrt{g+...}}}=12\\ \sqrt{g+12}=12\\ g+12=144\\ g=132\)

why was my post flagged lol