Sure! Let's factor the trinomial [tex]\( 2c^2 + 11c + 5 \)[/tex].
1. Identify the trinomial: We start with [tex]\( 2c^2 + 11c + 5 \)[/tex].
2. Find two numbers that multiply to the product of the coefficient of [tex]\( c^2 \)[/tex] ([tex]\(2\)[/tex]) and the constant term ([tex]\(5\)[/tex]). The product of these two numbers should be [tex]\( 2 \times 5 = 10 \)[/tex]. These two numbers should also add up to the middle coefficient ([tex]\(11\)[/tex]).
The two numbers that satisfy these conditions are [tex]\( 10 \)[/tex] and [tex]\( 1 \)[/tex] because:
[tex]\[
10 \times 1 = 10 \quad \text{and} \quad 10 + 1 = 11
\][/tex]
3. Rewrite the middle term using these two numbers: Instead of writing [tex]\( 11c \)[/tex], we write it as [tex]\( 10c + 1c \)[/tex]:
[tex]\[
2c^2 + 11c + 5 = 2c^2 + 10c + c + 5
\][/tex]
4. Factor by grouping: Group the terms in pairs and factor out the greatest common factor (GCF) from each pair:
[tex]\[
2c^2 + 10c + c + 5 = 2c(c + 5) + 1(c + 5)
\][/tex]
5. Factor out the common binomial factor: Notice that [tex]\( (c + 5) \)[/tex] is a common factor:
[tex]\[
2c(c + 5) + 1(c + 5) = (2c + 1)(c + 5)
\][/tex]
So the correct factorization of [tex]\( 2c^2 + 11c + 5 \)[/tex] is [tex]\((2c + 1)(c + 5)\)[/tex].
Therefore, the correct numbers to fill in the blanks in [tex]\( (2c + \_)(c + \_) \)[/tex] are:
- First blank: [tex]\(1\)[/tex]
- Second blank: [tex]\(5\)[/tex]
Hence, the factorization is [tex]\((2c + 1)(c + 5)\)[/tex].
