The question requires us to simplify the polynomials and then affix the right values in the table.
#1:
The polynomial is:
[tex](x-\frac{1}{2})(6x+2)[/tex]Let us multiply out the bracket and simplify:
[tex]\begin{gathered} (x-\frac{1}{2})(6x+2) \\ x(6x)+2x-\frac{1}{2}(6x)-2(\frac{1}{2}) \\ 6x^2+2x-3x-1 \\ \\ \therefore6x^2-x-1_{} \end{gathered}[/tex]The simplified form is given below:
[tex]6x^2-x-1[/tex]Its degree is quadratic.
There are 3 terms therefore, it is a trinomial
#2:
The polynomial is:
[tex]\begin{gathered} (7x^2+3x)-\frac{1}{3}(21x^2-12)\text{ (expand the brackets)} \\ 7x^2+3x-\frac{1}{3}(21x^2)+\frac{1}{3}(12) \\ \\ 7x^2+3x-7x^2+4 \\ \\ \therefore3x+4 \end{gathered}[/tex]The simplified form is: 3x + 4
Its degree is linear
There are 2 terms therefore, it is binomial
#3:
The polynomial is:
[tex]\begin{gathered} 4(5x^2-9x+7)+2(-10x^2+18x-13) \\ \exp and\text{ the brackets} \\ \\ 4(5x^2)-4(9x)+4(7)+2(-10x^2)+2(18x)+2(-13) \\ 20x^2-36x+28-20x^2_{}+36x-26 \\ \text{collect like terms} \\ \\ 20x^2-20x^2-36x+36x+28-26 \\ \\ \therefore2 \end{gathered}[/tex]The simplified form is: 2
Its degree is 0 i.e. a Constant
There is only 1 term, therefore, it is Monomial