Question

I need help with this

Answer 1

The question requires us to simplify the polynomials and then affix the right values in the table.

#1:

The polynomial is:

`(x-(1)/(2))(6x+2)`

Let us multiply out the bracket and simplify:

`\begin{gathered} (x-(1)/(2))(6x+2) \\ x(6x)+2x-(1)/(2)(6x)-2((1)/(2)) \\ 6x^2+2x-3x-1 \\ \\ \therefore6x^2-x-1_{} \end{gathered}`

The simplified form is given below:

`6x^2-x-1`

Its degree is quadratic.

There are 3 terms therefore, it is a trinomial

#2:

The polynomial is:

`\begin{gathered} (7x^2+3x)-(1)/(3)(21x^2-12)\text{ (expand the brackets)} \\ 7x^2+3x-(1)/(3)(21x^2)+(1)/(3)(12) \\ \\ 7x^2+3x-7x^2+4 \\ \\ \therefore3x+4 \end{gathered}`

The simplified form is: 3x + 4

Its degree is linear

There are 2 terms therefore, it is binomial

#3:

The polynomial is:

`\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}`

The simplified form is: 2

Its degree is 0 i.e. a Constant

There is only 1 term, therefore, it is Monomial