It has 1 sign change going from left to right. so it has 1 positive zero, which means it has either 2 or 0 negative zeros. The only rational zeros it might have are the fractions whose numerators divide evenly into the last term and whose denominators divide evenly into the first coefficient, 8. So the only possible rational zeros are ±1/1. ±1/2, ±1/4, ±1/8 or ±1. ±1/2, ±1/4, ±1/8 We try 1 1 | 8 -10 -7 -1 | 8 -2 -9 ------------- 8 -2 -9 -10 No luck. We try 1/2 1/2 | 8 -10 -7 -1 | 4 -3 -5 --------------- 8 -6 -10 -6 No luck. We try 1/4 1/4 | 8 -10 -7 -1 | 2 -2 -9/4 --------------- 8 -8 -9 -13/4 No luck. We try 1/8 1/8 | 8 -10 -7 -1 | 1 -9/8 -65/64 -------------------- 8 -9 -65/8 -129/64 No luck. We try -1 -1 | 8 -10 -7 -1 | -8 18 -11 -------------- 8 -18 11 -12 No luck. But, we learned that since the signs on that bottom row alternate and we are looking for a negative zero, that -1 is a lower bound for the negative zeros. But heck, the only possibilities of a rational zero are greater than -1 anyway. So, that didn't really help. We try -1/2 -1/2 | 8 -10 -7 -1 | -4 7 0 --------------- 8 -14 0 -1 No luck. We try -1/4 -1/4 | 8 -10 -7 -1 | -2 3 1 -------------- 8 -12 -4 0 Hey yeah! -1/4 is a zero So now we have factored the original polynomial equation this way: So now we factor the by taking out 4. Setting each factor = 0: we already knew that was one. , , So there are three zeros: , , and Edwin