How do you simplify (7i)^2?
374860
Well here is the math involved:\n\n(7i)^2 =  (7^2)(i^2) = 49(-1) = -49
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand
914360
You need to put parenthesis around your numerator if you mean that 7-13i is on top of the fraction.  You also are canceling the terms wrong.\n\nI am not sure what the problem is asking you to do.\n(7-13i)/(-2i)\nThat is already in its simplest form; the only thing you could do to it is make it read:\n-3.5/i + 6.5\nand that is more of a personal preference thing than a practical, useful thing.  If you are trying to find the value of i, you have to tell us what (7-13i)/(-2i) is equal to.
771836
Simple substitution.\n\n5(3) - 2(2) + 4(4)\n15 - 4 + 16\n27
285776
outside, put 4i  inside put 1  0  8  26i  -20\n\n4i)..1....0....8....26i....-20.\n.....1...4i...-8....-6i......4     \n\n(remember i^2 = -1)\n\n\n4i because 4i times 1 + 0 = 4i\n-8 because 4i times 4i is 16i^2 which is -16, plus 8 = -8\n-6i because 4i times -8 is -32i plus 26i = -6i\n4 because 4i times -6i is -24i^2 which is 24 minus 20 is 4\n\nSo the answer is x^3 + 4ix^2 - 8x - 6i, remainder 4
621340
The first one is (x-2)(x+9)(x-7).\nHowever, for the second part, everyone else has forgotten that imaginary zeroes come in CONJUGATE PAIRS.\nSo there are actually 5 zeroes to part 2, 4i, -5, -2i, 2i, -4i.\n\nSo the solution would be (x-4i)(x+4i)(x+5)(x-2i)(x+2i)\n(x^2+16)(x+5)(x^2+4)\nx^5 + 5x^4 + 20x^3 +100x^2 + 64x + 320\nViola...
337853
The answer, before simplification, is sqrt(-8). There is no need to panic, as some of the other answerers did. This simplifies easily to\n\n2 * i * sqrt(2)
1016236
(1/(z^2 - 3z - 4)) + ((3z)/(z^2 - 6z - 7))\n(1/((z - 4)(z + 1))) + ((3z)/((z - 7)(z + 1)))\n\nMultiply everything by (z - 4)(z + 1)(z - 7)\n\n((z - 7) + (3z(z - 4)))/((z - 4)(z + 1)(z - 7))\n\n(z - 7 + 3z^2 - 12z)/((z - 4)(z + 1)(z - 7))\n\n(3z^2 - 11z - 7)/((z - 4)(z + 1)(z - 7))\n\nANS : a.) (3z^2 - 11z - 7)/((z - 4)(z + 1)(z - 7))
742127
= (x^12)/12 - (8x^11)/11 + (5x^10)/2 - (10x^9)/3 - (15x^8)/4 + 24x^7 - 49x^6 + 60x^5 - (195x^4)/4 + (80x^3)/3 - (19x^2)/2 + 2x + C
1169872
(7/(3x)) / (x/3)\n(7/(3x)) * (3/x)\n21/(3x^2)\n7/x^2\n\nBut if you meant\n\n((7/3)x)/(x/3)\n((7x)/3)*(3/x)\n(21x)/(3x)\nANS : 7\n\n-------------------------------\n\n2 1/3 - 3/4\n2 1/3 = (7/3)\n\n(7/3) - (3/4)\nMultiply everything by 12\n(28 - 9)/12\n(19/12) or 1 (7/12)
544837
I squared the 7 to start with and got 49.  I am also only showing the addition version, you can do the subtraction later yourself.\n\n(-7 + sqrt(49 - 4(9)(-1)))/2(9)\n(-7 + sqrt(49 - -36))/18\n(-7 + sqrt(49 + 36))/18\n(-7 + sqrt(85))/18\n\nand I think that that is it.  You cannot simplify the square root.
954343
Use FOIL and multiply this out just like you would any other algebraic problem, and recall that i² = -1 when solving:\n\n(4-i)(5+i) = \n\n4*5 + 4i - 5i - i² = \n\n20 + 4i - 5i - (-1) =\n\n21 - i\n\nANSWER:  C. 21-i
201268
In doing this, you have to group the real parts together and the imaginary parts together. i.e.,\n(a + bi) + (c + di)\n= (a + c) + (bi + di) or\n= (a + c) + (b + d)i\n\nThe standard form of a complex number is:\na + bi, where a and b are real numbers\n\nThus,\n(5 + 3i) + (2 + 4i)\n= (5 + 2) + (3 + 4)i\n= 7 + 7i\n\nThe same goes with the second one, but with a slight twist:\ni - (6 + i) + (4 - 2i)\n= i - 6 - i + 4 - 2i\n= -6 + 4 + i - i - 2i\n= -2 - 2i\n\nIn multiplying, treat them as algebraic expression and remember that i² = -1\n\n-4i(3 - 7i)\n= -12i + 28i²\n= -12i + 28(-1)\n= -12i -28\n= -28 - 12i\n\nHope this helps&_&
825348
I prefer "long" multiplication but it doesn't line up as well...\n\n3x^2 + 7x - 6\n* _____x+12\n--------------------  Multiplying term by term\n_____36x^2 + 84X - 72\n3x^3 + 7x^2 - 6x\n--------------------------------  Add term by term\n3x^3 + 43x^2 + 78x - 72   <--  final answer
191203
(x-7i)(x+7i)=x^2 + (-i)^2 * 49 = x^2 + 49\n(which is your polinom having the roots -7i and +7i)
1121270
This one can't be simplified further uness you just write\nsqrt{5+3sqrt{3}}.\n\nOn the other hand, sqrt{7+4sqrt{3}} can be rewritten as 2+sqrt{3}. Making such simplifications is not easy in general.
140009
Note that it has two answers:\n\na^2=-4 ---> a=2i or a=-2i\n\n(2i)^2=2^2*i^2=4*(-1)=4\n\nand\n\n(-2i)^2=(-2)^2*i^2=4*(-1)=4