UnKoWn_AdventurE
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#1
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#1
one root of z^3 - 15z^ 2 + 76z - 140=0 is an integer. solve the equation
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ManLike007
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Through trial and error, I got z-7=0 as factor of the equation. Now use the long division method so that you're left with an equation in the form of (z-7)(az^2+bz+c)

Now you can't factorise the quadratic equation or use the quadratic formula because it doesn't satisfy b^2 - 4ac > 0 so you'll need to complete the square to find the two other roots.

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The roots I got:
z=7, z=4+2i and z=4-2i

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B_9710
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There is something called the rational roots theorem which says that if any rational roots exist to a polynomial then they are of the form  m/n where  m is a factor of 140 (constant term) and  n is a factor of 1 (coefficient of leading term). So the only values you should need to try are  \pm 1, 2, 4, 5, 7, 10, 20, 28, 35, 70, 140. If any rational roots exist then they are one of these 22 values. Of course you find that 7 works quickly if you use a calculator.
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