GCE 200 1) Given the polynomial p(x)=(x-2)(2x-1)(x+1) Solve the equation p(x)=0 GCE 202 Find the remainder when p(x) is divided by (x-1)

GCE 200
1) To solve the equation $$p(x)=0$$ , we need to find the values of $$x$$ that make the polynomial equal to zero. We can do this by setting each factor of the polynomial equal to zero and solving for $$x$$ .

$$(x-2)=0 \Rightarrow x=2$$
$$(2x-1)=0 \Rightarrow x=\frac{1}{2}$$
$$(x+1)=0 \Rightarrow x=-1$$

So, the solutions to the equation $$p(x)=0$$ are $$x=2$$ , $$x=\frac{1}{2}$$ , and $$x=-1$$ .

GCE 202
To find the remainder when $$p(x)$$ is divided by $$(x-1)$$ , we can use the Remainder Theorem. According to the Remainder Theorem, if a polynomial $$p(x)$$ is divided by $$(x-a)$$ , the remainder is equal to $$p(a)$$ .

In this case, we need to find the remainder when $$p(x)$$ is divided by $$(x-1)$$ , so we need to evaluate $$p(1)$$ .

$$p(1)=(1-2)(2(1)-1)(1+1)=(-1)(1)(2)=-2$$

Therefore, the remainder when $$p(x)$$ is divided by $$(x-1)$$ is $$-2$$ .