Problems from Remainder theorem and Factor theorem

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:

$p(1)=1^2-2\cdot 1+1=0$

$p(-1)=(-1{)}^2+2\cdot 1+1=4$

$p(2)=2^2-2\cdot 2+1=1$

$p(-2)=(-2{)}^2+2\cdot 2+1=9$

$q(1)=1^3-2\cdot 1^2-4\cdot 1+8=3$

$q(-1)=(-1{)}^3-2\cdot (-1{)}^2-4\cdot (-1)+8=-1-2+4+8=9$

$q(2)=2^3-2\cdot 2^2-4\cdot 2+8=0$

$q(-2)=(-2{)}^3-2\cdot (-2{)}^2-4\cdot (-2)+8=-8-8+8+8=0$

$r(1)=2$

$r(-1)=2$

$r(2)=5$

$r(-2)=5$

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.

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