** Example 1:** Find the remainder when x

^{4}+ x

^{3}– 2x

^{2}+ x + 1 is divided by x – 1.

Solution : Here, p(x) = x^{4} + x^{3} – 2x^{2} + x + 1, and the zero of x – 1 is 1.

So, p(1) = (1)^{4 }+ (1)^{3} – 2(1)^{2} + 1 + 1

= 2

So, by the Remainder Theorem, 2 is the remainder when x^{4} + x^{3} – 2x^{2} + x + 1 is divided by x – 1.

** Example 2:** Check whether the polynomial q(t) = 4t

^{3}+ 4t

^{2}– t – 1 is a multiple of 2t + 1.

Solution : As you know, q(t) will be a multiple of 2t + 1 only, if 2t + 1 divides q(t) leaving remainder zero. Now, taking 2t + 1 = 0, we have t = –^{1 }/_{2}.

Also,

q(-^{1 }/_{2}) = 4(-^{1 }/_{2}) ^{3} + 4(-^{1 }/_{2}) ^{2}– (-^{1 }/_{2}) – 1

= –^{1 }/_{2 }+ 1 + ^{1 }/_{2} – 1

= 0

So the remainder obtained on dividing q(t) by 2t + 1 is 0.

So, 2t + 1 is a factor of the given polynomial q(t), that is q(t) is a multiple of 2t + 1.

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