Find value of k in  (x+1)(x^2+kx-4)=x^3+4x^2-x-4?
963577
4x^2 is the key.\n\nYou get 1 x^2 by multiplying 1 by x^2, so kx times x must equal 4-1=3.\n\nIn the same way for the second:\n\nYou get -1 x^2 by multiplying -1 by x, so 2*k*x^2 = -5-(-1)x^2\n\n2*k*x^2 = -4x^2\nk=-4/2\nk=-2.\n\nThe second is similar.  Basically, look at the x^2 coefficient and look at how many you have through the terms NOT including k, then subtract the number of other ones from the total number of x^2.  kx must be multiplied by x to get x^2, so take into account the other x coefficient.\n\ni.e. (2x)(kx) = 2kx^2
1053464
if you want to simplify it \n\nit is 3x - 4x + 4 = -x + 4
122155
(x + 1)(x + 3)(x - 2) = 0\n(x^2 + 4x + 3)(x - 2) = 0\n(x^2)(x - 2) + 4x(x - 2) + 3(x - 2) = 0\nx^3 - 2x^2 + 4x^2 - 8x + 3x - 6 = 0\nx^3 + 2x^2 - 5x - 6 = 0 is the answer ;-)
1021313
Replace x with 6\n6^2 - 2(6) + k = 0\n36 - 12 + k = 0\n24 + k = 0\nk = -24\nThe other root is -24/6 = -4
846663
(x+3)(x-1)\n\n x= -3\nx=1
798648
(x+3)(x^2+2x+4)\n\nStart by taking the first term in the first parenthesis and multiplying it by each term in the second parenthesis:\n\nx*x^2 + x*2x + x*4 = x^3 + 2x^2 + 4x\n\nNow take the second term and multiply it by each term in the second parenthesis:\n\n3*x^2 + 3*2x + 3*4 = 3x^2 + 6x +12\n\nNow add all of that together\n\nx^3 + 2x^2 + 4x + 3x^2 + 6x +12\n\nand combine terms\n\nx^3 + 5x^2 + 10x +12
879176
= x(x+3) + 4(x+3)\n= x^2 + 3x +4x + 12\n= x^2 +7x + 12
853794
Evaluate the expression after the ∑ for each of k = 1, 2, ..., up to 6, and add them:\n\n[ 1 + (-1)^1 ] 2^-1 +\n[ 1 + (-1)^2 ] 2^-2 +\n[ 1 + (-1)^3 ] 2^-3 +\n[ 1 + (-1)^4 ] 2^-4 +\n[ 1 + (-1)^5 ] 2^-5 +\n[ 1 + (-1)^6 ] 2^-6\n\nNow use (-1)^1 = -1, (-1)^2 = (-1) × (-1) = 1, ... to get\n\n[ 1 + (-1) ] 2^-1 +\n[ 1 + (1) ] 2^-2 +\n[ 1 + (-1) ] 2^-3 +\n[ 1 + (1) ] 2^-4 +\n[ 1 + (-1) ] 2^-5 +\n[ 1 + (1) ] 2^-6\n\n=\n\n[ 0 ] 2^-1 +\n[ 2 ] 2^-2 +\n[ 0 ] 2^-3 +\n[ 2 ] 2^-4 +\n[ 0 ] 2^-5 +\n[ 2 ] 2^-6\n\n= 2 × 2^-2 + 2 × 2^-4 + 2 × 2^-6\n\nUsing a^m × a^n = a^(m + n), simplify the exponents, to get\n\n2^-1 + 2^-3 + 2^-5 \n= 1 / 2 + 1 / 8 + 1 / 32\n= 16 / 32 + 4 / 32 + 1 / 32 \n= 21 / 32
863644
take out 4x first\n4x(x2-6x+8)\n4x(x-4)(x-2)
665603
you put (x+2) in for x in the equation first\n\n((x+2)^3)+(2(x+2)^2)-4\n\nthen simplify \n\nx^3 + 8x^2 + 20x + 12\n\nand multiply the entire thing by 3\n\n3(x^3 + 8x^2 + 20x + 12)\n\n3x^3 + 24x^2 + 60x + 36
926616
2x+3<6x-1\n2x-6x<-1-3 \n-4x<-4\nx>1
785246
x^2-16=0\n(x+4)(x-4)=0\nx+4=0\nx=-4\n\nx-4=0\nx=-4\n\nx^3+14x^2+40x=0\nx(x^2-14x+40)=0\nx(x-10)(x-4)=0\nx=0\n\nx-10=0\nx=10\n\nx-4=0\nx=4\n\nx^4-5x^2+4=0\n(x^2-4)(x^2-1)=0\n(x-2)(x+2)(x-1)(x+1)=0\nxi2=0\nx=2\nx+2=0\nx=-2\nx-1=0\nx=1\nx+1=0\nx=-1
695176
6x^2-3x-4x+2\n3x(2x-1)-2(2x-1)\n(2x-1)(3x-2)
71279
It'sa simple aquation...\n\nx-4/4-x = x-1-x = x-x-1 = 0-1 = -1
720161
Yes, I can:\n4! + 4 \n=\n(4 x 3 x 2 x 1) + 4 = 28\n\n(BTW, 4squared + 4squared = 32)
817835
x^3-x^2+3x-1\nx+4&#92;x^4+3x^3-x^2+11x-4\n      -(x^4+4x^3)\n              -x^3-x^2\n            -(-x^3-4x^2)\n                       3x^2+11x\n                     -(3x^2+12x)\n                                   -x-4\n                                 -(-x-4)\n                                         0\nx^4+3x^3-x^2+11x-4=(x+4)(x^3-x^2+3x-1)
899666
2k + 4 + 2k = -6 + 3\n\n4k + 4 = -3\n\n4k = -7\n\nk = -7 / 4