Question 821259
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Hi,
Find the real zeroes of 
g(x)=x^3-11x^2+26x-16    |Note:  x = 1 is a root  (1-11 +26- 16 )= 0
Using synthetic division: dividing by (x-1)
1	1	-11	26	-16
		1	-10	16
	1	-10	16	0
g(x) =  (x-1)(x^2 - 10x + 16)
g(x) =  (x-1)(x - 8)(x - 2)
 roots are:  1, 2 , 8