Answered

Drag each label to the correct location on the table. Each label can be used more than once, but not all labels will be used.

Simplify the given polynomials. Then, classify each by its degree and number of terms.

[tex]\[
\begin{array}{ll}
\text { Polynomial 1: } & \left(3 x-\frac{1}{4}\right)(4 x+8) \\
\text { Polynomial 2: } & \left(5 x^2+7 x\right)-\frac{1}{2}\left(10 x^2-4\right) \\
\text { Polynomial 3: } & 3\left(8 x^2+4 x-2\right)+6\left(-4 x^2-2 x+3\right)
\end{array}
\][/tex]

\begin{tabular}{|c|c|c|c|}
\hline Polynomial & Simplified Form & \begin{tabular}{c}
Name by \\
Degree
\end{tabular} & \begin{tabular}{c}
Name by \\
Number of Terms
\end{tabular} \\
\hline Polynomial 1 & & & trinomial \\
\hline Polynomial 2 & & linear & \\
\hline Polynomial 3 & 12 & & \\
\hline
\end{tabular}

Labels:
- quadratic
- [tex]\(3 x^2 + 4 x - 2\)[/tex]
- constant
- [tex]\(12 x^2 + 23 x - 2\)[/tex]
- monomial
- [tex]\(7 x + 2\)[/tex]
- [tex]\(12 x + 20\)[/tex]
- binomial



Answer :

GinnyAnswer
Let's simplify each polynomial step-by-step and then classify them by degree and number of terms.

### Polynomial 1: [tex]\(\left(3x - \frac{1}{4}\right)(4x + 8)\)[/tex]
To simplify this, use the distributive property (also known as the FOIL method for binomials):
[tex]\[ (3x - \frac{1}{4})(4x + 8) = 3x \cdot 4x + 3x \cdot 8 - \frac{1}{4} \cdot 4x - \frac{1}{4} \cdot 8 \][/tex]
[tex]\[ = 12x^2 + 24x - x - 2 \][/tex]
Combine like terms:
[tex]\[ = 12x^2 + 23x - 2 \][/tex]
This polynomial is a quadratic trinomial because its highest degree is 2 (quadratic) and it has three terms (trinomial).

### Polynomial 2: [tex]\(\left(5x^2 + 7x\right) - \frac{1}{2}\left(10x^2 - 4\right)\)[/tex]
Simplify the expression inside the parentheses and distribute the [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ 5x^2 + 7x - \frac{1}{2}(10x^2 - 4) = 5x^2 + 7x - 5x^2 + 2 \][/tex]
Combine like terms:
[tex]\[ = 7x + 2 \][/tex]
This polynomial is a linear binomial because its highest degree is 1 (linear) and it has two terms (binomial).

### Polynomial 3: [tex]\(3\left(8x^2 + 4x - 2\right) + 6\left(-4x^2 - 2x + 3\right)\)[/tex]
Distribute the constants inside the parentheses:
[tex]\[ 3(8x^2 + 4x - 2) + 6(-4x^2 - 2x + 3) = 24x^2 + 12x - 6 - 24x^2 - 12x + 18 \][/tex]
Combine like terms:
[tex]\[ = (24x^2 - 24x^2) + (12x - 12x) + (-6 + 18) = 0x^2 + 0x + 12 = 12 \][/tex]
This polynomial is a constant monomial because it has no variable terms (constant) and it has one term (monomial).

Now we place the simplified forms and classifications in the table:

[tex]\[ \begin{array}{|c|c|c|c|} \hline \text{Polynomial} & \text{Simplified Form} & \text{Name by Degree} & \text{Name by Number of Terms} \\ \hline \text{Polynomial 1} & 12x^2 + 23x - 2 & \text{quadratic} & \text{trinomial} \\ \hline \text{Polynomial 2} & 7x + 2 & \text{linear} & \text{binomial} \\ \hline \text{Polynomial 3} & 12 & \text{constant} & \text{monomial} \\ \hline \end{array} \][/tex]

This is the completed table:
[tex]\[ \begin{array}{|c|c|c|c|} \hline \text{Polynomial} & \text{Simplified Form} & \text{Name by Degree} & \text{Name by Number of Terms} \\ \hline \text{Polynomial 1} & 12x^2 + 23x - 2 & \text{quadratic} & \text{trinomial} \\ \hline \text{Polynomial 2} & 7x + 2 & \text{linear} & \text{binomial} \\ \hline \text{Polynomial 3} & 12 & \text{constant} & \text{monomial} \\ \hline \end{array} \][/tex]