find all real and imaginary zeros
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When I see this sort of question, I make a note of the coefficients: 1 10 23 14 and I add them together to see if the y add up to zero. If the do, I know that x=1 is a solution. Now obviously they don't in this case, so I look for the odd powers of x and make the coefficients of these negative: -1 10 -23 14. Again, I add them up, and if they come to zero I know that x=-1 is a solution. In this case, it is. Now I use synthetic division to divide by this solution:

-1 | 1 10 23..14

......1 -1..-9 -14

......1..9 14 | 0

Now I only have a quadratic to solve: x^2+9x+14=(x+2)(x+7), so the complete factorisation is (x+1)(x+2)(x+7) and the zeroes are -1, -2, -7.

by Top Rated User (1.1m points)

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