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

\(p\left( -\frac { m }{ l } \right) =l\left( -\frac { m }{ l } \right) +m=-m+m=0\)
\(\therefore -\frac { m }{ l } \) is a zero of p(x)

It is a non - zero constant. So the degree of this polynomial is zero.

Let \(f(x)={ x }^{ 4 }+{ x }^{ 3 }-7{ x }^{ 2 }-x+6\)
By trail, f(1)=0 and f(2)=0
So by factor theorem, (x-1) and (x-2) are factors of f(x).
\((x-1)(x-2)={ x }^{ 2 }-3x+2\)
Now \(f(x)={ x }^{ 2 }({ x }^{ 2 }-3x+2)+4x({ x }^{ 2 }-3x+2)+3({ x }^{ 2 }-3x+2)\)
\(=({ x }^{ 2 }-3x+2)({ x }^{ 2 }+4x+3)\)
\(=(x-1)(x-2)(x+1)(x+3)\)
\({ x }^{ 2 }+4x+3\)
\(={ x }^{ 2 }+c+3x+3\)
\(={ x }^{ 2 }+c+3x+3\)
\(=(x+1)(x+3)\)

Number of terms = 3
Terms: 3x², -5x, 7

0

monomial

5

The terms containing x² is-x² , i.e., (-1) x²

\((x^3+5)(4-x^5)\)\(=4x^3-x^8+20-5x^5\)

Fractional power of x in (b),
Two variables in (c),
Negative power of x in (d)