Sure, let's factor the trinomial [tex]\(x^2 - 9x + 20\)[/tex].
To factor the trinomial of the form [tex]\(ax^2 + bx + c\)[/tex], we are looking for two binomials [tex]\((x - m)(x - n)\)[/tex] such that:
1. [tex]\(m \cdot n = c\)[/tex]
2. [tex]\(m + n = -b\)[/tex]
Given the trinomial [tex]\(x^2 - 9x + 20\)[/tex]:
1. The constant term [tex]\(c = 20\)[/tex].
2. The coefficient of [tex]\(x\)[/tex] is [tex]\(-9\)[/tex], thus [tex]\(b = -9\)[/tex].
We need two numbers, [tex]\(m\)[/tex] and [tex]\(n\)[/tex], that multiply to 20 and add up to -9.
After analyzing the factors of 20, we identify that:
- [tex]\(5 \cdot 4 = 20\)[/tex]
- [tex]\(5 + 4 = 9\)[/tex], but we need -9, so we use -5 and -4.
Thus, the factors of the trinomial [tex]\(x^2 - 9x + 20\)[/tex] are:
[tex]\[(x - 5)(x - 4)\][/tex]
So, the correct answer to the question is:
([tex]$x$[/tex] [tex]$-$[/tex] [tex]$5$[/tex]) ([tex]$x$[/tex] [tex]$-$[/tex] [tex]$4$[/tex]).
