(x+1)(x+5)(x+2)
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x(x+3)(x+5)
x3 + x2 - 17x + 15 = (x - 1)(x - 3)(x + 5). Thus, the zeros are 1, 3, and -5. All three zeros are rational.
Assumption 8x2 = eight times x squared Assumption: x3 is x to the third Assumption: y2 is y squared Answer 40 y to the fourth x to the fifth = 40 y4 x5
x3 + 7x = 8x2x3 + 7x - 8x2 = 0 [subtract 8x2 from both sides]x(x2 + 7 - 8x) = 0 [factor out x]x(x-1)(x-7) = 0 [factor]Since the product of the three factors x, x-1, and x-7 equals zero, any of the three expressions could equal zero:x = 0x - 1 = 0, x = 1x - 7 = 0, x = 7Therefore, there are three solutions to the equation x3 + 7x = 8x2:x = 0x = 1x = 7Or, in set notation: x = {0,1,7}
x(x - 17)(x - 1)Factorising x3 - 18x2 + 17x:Common factor x of all terms:x(x2 - 18x + 17)17 = -1 x -17, -1 + -17 = -18:x(x - 1)(x - 17)