\(x^3+13x^2+32x+20\)
Let \(p(x)={ x }^{ 3 }+13{ x }^{ 2 }+32x+20\)
By trail, we find that
\(p(-1)={ (-1) }^{ 3 }+13{ (-1) }^{ 2 }+32(-1)+20\)
\(=-1+13-32+20=0\)
\(\therefore\) By Factor Theorem, x-(-1), i.e., (x+1) is a factor of p(x).
Now,
\({ x }^{ 3 }+13{ x }^{ 2 }+32x+20={ x }^{ 2 }(x+1)+12x(x+1)+20(x+1)\)
\(=(x+1)({ x }^{ 2 }+12x+20)\)
\(=(x+1)({ x }^{ 2 }+2x+10x+20)\)
\(=(x+1)\{ x(x+2)+10(x+2)\} \)
\(=(x+1)(x+2)(x+10).\)

Let \(p(x)={ x }^{ 3 }-2{ x }^{ 2 }-x+2\)
By trail, we find that
\(p(1)={ (1) }^{ 3 }-2{ (1) }^{ 2 }-(1)+2\)
\(=1-2-1+2=0\)
\(\therefore \) By Factor Theorem, (x-1) is a factor of p(x).
Now,
\({ x }^{ 3 }-2{ x }^{ 2 }-x+2={ x }^{ 2 }(x-1)-x(x-1)-2(x-1)\)
\(=(x-1)({ x }^{ 2 }-x-2)\)
\(=(x-1)({ x }^{ 2 }-2x+x-2)\)
\(=(x-1)\{ x(x-2)+1(x-2)\} \)
\(=(x-1)(x-2)(x+1).\)

\({ 3 }x^{ 2 }-x-4\)
\({ 3 }x^{ 2 }-x-4={ 3 }x^{ 2 }-4x+3x-4\)
\(=x(3x-4)+1(3x-4)\)
\(=(3x-4)(x+1)\)

\({ x }^{ 3 }-{ x }^{ 2 }-(2+\sqrt { 2 } )x+\sqrt { 2 } .\)
Let  \(p(x)=\)\({ x }^{ 3 }-{ x }^{ 2 }-(2+\sqrt { 2 } )x+\sqrt { 2 } .\)
The zero of x+1 is -1
\(​p(-1)=(-1)^{ 3 }+3(-1)^{ 2 }+(2-\sqrt { 2 } )(-1)+\sqrt { 2 }\)
\( \\ =-1-1+2+\sqrt { 2 } +\sqrt { 2 } =2\sqrt { 2 } \neq 0\)
\(\therefore\) By factors theorem, x+1 is not a factor of \({ x }^{ 3 }-{ x }^{ 2 }-(2+\sqrt { 2 } )x+\sqrt { 2 } .\)

\(x^4+3x^3+3x^2+x+1\)
Let \(p(x)=\)\(x^4+3x^3+3x^2+x+1\)
The zero of x+1 is -1.
\(​p(-1)=(-1)^4+3(-1)^3+(3-1)^2+(-1)+1\)
\(\\ =1-3+3-1+1=1\neq 0\)
\(\therefore\) By factors theorem, x+1 is not a factor of \(x^4+3x^3+3x^2+x+1\)

\(x+\pi\)
\(x+\pi=0\ d\Rightarrow\ x=-\pi\)
\(\therefore\) Remainder \(= (-\pi)^3+3(-\pi)^2+3(-\pi)+1\)
 \(= -\pi^3+3\pi^2+3\pi+1\)

\(x\)
\(\therefore\) Remainder
 \(= (0)^3+3(0)^2+3(0)+1=1\)

\(p\left( -\frac { 1 }{ \sqrt { 3 } } \right) =3\left( \frac { 1 }{ \sqrt { 3 } } \right) ^{ 2 }-1\)
\(\\ \ =3\left( \frac { 1 }{ 3 } \right) -1=1-1=0\)
\(\\ p\left( \frac { 2 }{ \sqrt { 3 } } \right) =3\left( \frac { 2 }{ \sqrt { 3 } } \right) ^{ 2 }-1=3\left( \frac { 4 }{ 3 } \right) -1\)
\(\\ \ =4-1=3\neq 0\)
\(\therefore -\frac { 1 }{ \sqrt { 3 } } \) is a zero of p(x) but \(\frac { 2 }{ \sqrt { 3 } } \) is not a zero of p(x)

\(\therefore \ p(0)=(0-1)(0+1)=(-1)(1)=-1\)
\(\\ p(1)=(1-1)(1+1)=(0)(2)=0\)
and \(p(2)=(2-1)(2+1)=(1)(3)=3\)

\(\therefore \ p(0)={ (0) }^{ 3 }=0,\)
\(p(1)={ (1) }^{ 3 }=1\)
and   \(p(2)={ (2) }^{ 3 }=8\)

quadratic

Term with the highest power of x = 5x³
Exponent of x in this term           = 3
Therefore Degree of this polynomial = 3

Coefficient of x² = 1

This expression is a polynomial in one variable x because in the expression there is only one variable (x) and all the indices of x are whole numbers.