Verify, without finding explicitly the roots, which of these values are roots of the following polynomials:
values: $$1,-1,2,-2$$
polynomials:
$$p(x)=x^2-2x+1$$
$$q(x)=x^3-2x^2-4x+8$$
$$r(x)=x^2+1$$
Development:
A necessary and sufficient condition so that a value $$a$$ is a root of a polynomial $$p(x)$$, is that $$p(a)=0$$. Then, we keep trying with the different values for each polynomial:
$$p(1)=1^2-2\cdot1+1=0$$
$$p(-1)=(-1)^2+2\cdot1+1=4$$
$$p(2)=2^2-2\cdot2+1=1$$
$$p(-2)=(-2)^2+2\cdot2+1=9$$
$$q(1)=1^3-2\cdot1^2-4\cdot1+8=3$$
$$q(-1)=(-1)^3-2\cdot(-1)^2-4\cdot(-1)+8=-1-2+4+8=9$$
$$q(2)=2^3-2\cdot2^2-4\cdot2+8=0$$
$$q(-2)=(-2)^3-2\cdot(-2)^2-4\cdot(-2)+8=-8-8+8+8=0$$
$$r(1)=2$$
$$r(-1)=2$$
$$r(2)=5$$
$$r(-2)=5$$
Solution:
$$p(x)$$ has $$1$$ as a root.
$$q(x)$$ has $$2$$ and $$-2$$ as a root.
$$r(x)$$ does not have any of the values in the list as a root. In fact, it does not have any rational value as a root, therefore it is an irreducible polynomial.