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16x2 - 128x + 256 = 16(x2 - 8x + 16) = 16(x - 4)2, or [4(x - 4)]2, or (4x - 16)2
5x-6+3x+128x-6+128x+6
To be rational, the numerator has to be a factor of 52 (the last term), and the denominator has to be a factor of 3 (the coefficient of the first term). Try all the combinations, including negative numbers (52/1, -52/1, 52/3, -52/3, 26/1, 26/3, etc.), to see whether one of them is a zero.
11 = (42 - 4) - (4 / 4) 12 = (4 + 4) + (√4 + √4) 13 = (42 - 4) + (4 / 4) 14 = (4 + 4 + 4 + √4) 15 = (4 * 4) - (4 / 4) 16 = (4 + 4 + 4 + 4) 17 = (42 + √4) - (4 / 4) 18 = (42 + 4) - (4 - √4) 19 = (42 + 4) - (4 / 4) 20 = (4 * 4) + (√4 + √4)
Here is one set of solutions. The answers here are not unique. 1 = (4*4)/(4*4) 2 = 4/4 + 4/4 3 = (4+4+4)/4 4 = (4-4)*4 + 4 5 = (4*4 + 4) / 4 6 = 4 + (4+4)/4 7 = 4 + 4 - 4/4 8 = 4 + 4 + 4 - 4 9 = 4 + 4 + 4/4 10 = (44 - 4)/4