Sure! Let's factor the given quadratic expression completely by following a step-by-step approach.
Given quadratic expression:
[tex]\[ x^2 + 11x + 30 \][/tex]
### Step 1: Identify the coefficients
The quadratic expression is in the form [tex]\( ax^2 + bx + c \)[/tex], where:
- [tex]\( a = 1 \)[/tex] (coefficient of [tex]\( x^2 \)[/tex])
- [tex]\( b = 11 \)[/tex] (coefficient of [tex]\( x \)[/tex])
- [tex]\( c = 30 \)[/tex] (constant term)
### Step 2: Find two numbers that multiply to [tex]\( a \times c \)[/tex] and add to [tex]\( b \)[/tex]
We need to find two numbers that:
- Multiply to [tex]\( a \times c = 1 \times 30 = 30 \)[/tex]
- Add to [tex]\( b = 11 \)[/tex]
The pairs of factors of 30 are:
- [tex]\( 1 \times 30 \)[/tex]
- [tex]\( 2 \times 15 \)[/tex]
- [tex]\( 3 \times 10 \)[/tex]
- [tex]\( 5 \times 6 \)[/tex]
Among these pairs, the pair that adds up to 11 is [tex]\( 5 \)[/tex] and [tex]\( 6 \)[/tex].
### Step 3: Express the middle term as the sum of these two numbers
We rewrite [tex]\( 11x \)[/tex] as [tex]\( 5x + 6x \)[/tex]:
[tex]\[ x^2 + 11x + 30 = x^2 + 5x + 6x + 30 \][/tex]
### Step 4: Factor by grouping
Group the terms to factor by grouping:
[tex]\[ x^2 + 5x + 6x + 30 = (x^2 + 5x) + (6x + 30) \][/tex]
Factor out the common factors in each group:
[tex]\[ (x^2 + 5x) + (6x + 30) = x(x + 5) + 6(x + 5) \][/tex]
### Step 5: Factor out the common binomial factor
Notice that [tex]\((x + 5)\)[/tex] is a common factor in both terms:
[tex]\[ x(x + 5) + 6(x + 5) = (x + 5)(x + 6) \][/tex]
### Final Answer
The given quadratic expression [tex]\( x^2 + 11x + 30 \)[/tex] factors completely as:
[tex]\[ (x + 5)(x + 6) \][/tex]
And that's the fully factored form of the quadratic expression!
