1.

Show that (x – 3) is a factor of x3 + 9x2 – x – 105.

Answer»

Let p(x) = x3 + 9x2 – x – 105 

By factor theorem, x – 3 is a factor of p(x), if p(3) = 0 

p(3) = 33 + 9(3)3 – 3 – 105

= 27 + 81 – 3 – 105

 = 108 – 108 

p(3) = 0

To find the zero of x - 3:

Put x - 3 = 0 

We get x = 3

Therefore, x – 3 is a factor of x3 + 9x2 – x – 105.



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