(i) Let p(x)=x3–x2+11x+69
We have to show that, x + 3 is a factor of p(x).
i.e. p(-3) = 0 [Using Remainder Theorem]
Now, p(−3)=(−3)3–(−3)2+11(−3)+69
=−27–9–33+69=−69+69=0.
Therefore, x +3 is a factor.
(ii) p(x)=2x3−9x2+x+12
We have to show that, 2x - 3 is a factor of p(x).
i.e. p(32)=0 [Using Remainder Theorem]
p(32)=2(32)3−9(32)2+32+12
=274−814+32+12
=81−814=0
Hence, (2x-3) is a factor of p(x).