Let's solve for the correct pairs of trinomials and their factors. We need to match the given trinomials with the correct factors.
Here are the given trinomials:
1. [tex]\( a^2 - 12a + 20 \)[/tex]
2. [tex]\( a^2 - 19a - 20 \)[/tex]
And here are the possible factors:
1. [tex]\( (a-2)(a-10) \)[/tex]
2. [tex]\( (a-4)(a+5) \)[/tex]
3. [tex]\( (a-20)(a+1) \)[/tex]
Matching Trinomials with Factors:
1. For the trinomial [tex]\( a^2 - 12a + 20 \)[/tex]:
To factorize this trinomial, we look for two numbers that multiply to give the constant term (20) and add to give the coefficient of the middle term (-12).
We find that [tex]\( (a - 2)(a - 10) \)[/tex] works since:
- [tex]\( (a - 2) \times (a - 10) = a^2 - 10a - 2a + 20 = a^2 - 12a + 20 \)[/tex]
Thus, [tex]\( a^2 - 12a + 20 \)[/tex] factors as [tex]\( (a - 2)(a - 10) \)[/tex].
2. For the trinomial [tex]\( a^2 - 19a - 20 \)[/tex]:
To factorize this trinomial, we look for two numbers that multiply to give the constant term (-20) and add to give the coefficient of the middle term (-19).
We find that [tex]\( (a - 20)(a + 1) \)[/tex] works since:
- [tex]\( (a - 20) \times (a + 1) = a^2 + a - 20a - 20 = a^2 - 19a - 20 \)[/tex]
Thus, [tex]\( a^2 - 19a - 20 \)[/tex] factors as [tex]\( (a - 20)(a + 1) \)[/tex].
Therefore, the correct pairs are:
[tex]\[
\begin{array}{ccc}
\text{Trinomial} & & \text{Factor} \\
\hline
a^2 - 12a + 20 & \longrightarrow & (a-2)(a-10) \\
a^2 - 19a - 20 & \longrightarrow & (a-20)(a+1) \\
\end{array}
\][/tex]
