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To factor the quadratic trinomial [tex]\(x^2 + 6x + n = 0\)[/tex], we need to find two numbers that multiply to [tex]\(n\)[/tex] and add up to the coefficient of [tex]\(x\)[/tex], which is 6.
Given the equation:
[tex]\[ x^2 + 6x + n = 0 \][/tex]
Here is the step-by-step process:
1. Understand the form we need:
We need to factor it into the form:
[tex]\[ (x + p)(x + q) = 0 \][/tex]
where [tex]\(p\)[/tex] and [tex]\(q\)[/tex] are the numbers that we are looking for.
2. Set up the equations based on the form:
Expanding the factored form [tex]\( (x + p)(x + q) \)[/tex]:
[tex]\[ x^2 + (p + q)x + pq = 0 \][/tex]
By comparing coefficients with [tex]\(x^2 + 6x + n = 0\)[/tex]:
[tex]\[ p + q = 6 \quad \text{(1)} \][/tex]
[tex]\[ pq = n \quad \text{(2)} \][/tex]
3. Solving the system of equations:
We need to find pairs [tex]\((p, q)\)[/tex] that satisfy both equations simultaneously.
4. Example for a specific [tex]\(n\)[/tex]:
For this step, let’s consider different possible values of [tex]\(n\)[/tex] to illustrate the method.
- Case 1: [tex]\(n = 8\)[/tex]
- We need to find [tex]\(p\)[/tex] and [tex]\(q\)[/tex] such that:
[tex]\[ pq = 8 \quad \text{and} \quad p + q = 6 \][/tex]
- The pairs that satisfy these equations are [tex]\(p = 2\)[/tex] and [tex]\(q = 4\)[/tex].
- Thus, the factored form is:
[tex]\[ (x + 2)(x + 4) = 0 \][/tex]
- Case 2: [tex]\(n = 9\)[/tex]
- Here, we need [tex]\(pq = 9\)[/tex] and [tex]\(p + q = 6\)[/tex].
- The pairs that satisfy these equations are [tex]\(p = 3\)[/tex] and [tex]\(q = 3\)[/tex].
- Thus, the factored form is:
[tex]\[ (x + 3)(x + 3) = 0 \quad \text{or} \quad (x + 3)^2 = 0 \][/tex]
- Case 3: [tex]\(n = 7\)[/tex]
- Now, we need [tex]\(pq = 7\)[/tex] and [tex]\(p + q = 6\)[/tex].
- Unfortunately, there are no integer pairs for [tex]\(p\)[/tex] and [tex]\(q\)[/tex] that satisfy these equations.
5. Factoring in general:
Often [tex]\(n\)[/tex] is known or can typically be simplified further. While in our cases [tex]\(n\)[/tex] was straightforward enough for direct factoring.
In conclusion, solving [tex]\(x^2 + 6x + n = 0\)[/tex] for different [tex]\(n\)[/tex] requires identifying pairs [tex]\((p, q)\)[/tex] that add up to the middle coefficient (in this case, 6) and multiply to the constant term [tex]\(n\)[/tex]. Depending on [tex]\(n\)[/tex], if integer solutions exist, factoring can be manually achieved as shown.
Given the equation:
[tex]\[ x^2 + 6x + n = 0 \][/tex]
Here is the step-by-step process:
1. Understand the form we need:
We need to factor it into the form:
[tex]\[ (x + p)(x + q) = 0 \][/tex]
where [tex]\(p\)[/tex] and [tex]\(q\)[/tex] are the numbers that we are looking for.
2. Set up the equations based on the form:
Expanding the factored form [tex]\( (x + p)(x + q) \)[/tex]:
[tex]\[ x^2 + (p + q)x + pq = 0 \][/tex]
By comparing coefficients with [tex]\(x^2 + 6x + n = 0\)[/tex]:
[tex]\[ p + q = 6 \quad \text{(1)} \][/tex]
[tex]\[ pq = n \quad \text{(2)} \][/tex]
3. Solving the system of equations:
We need to find pairs [tex]\((p, q)\)[/tex] that satisfy both equations simultaneously.
4. Example for a specific [tex]\(n\)[/tex]:
For this step, let’s consider different possible values of [tex]\(n\)[/tex] to illustrate the method.
- Case 1: [tex]\(n = 8\)[/tex]
- We need to find [tex]\(p\)[/tex] and [tex]\(q\)[/tex] such that:
[tex]\[ pq = 8 \quad \text{and} \quad p + q = 6 \][/tex]
- The pairs that satisfy these equations are [tex]\(p = 2\)[/tex] and [tex]\(q = 4\)[/tex].
- Thus, the factored form is:
[tex]\[ (x + 2)(x + 4) = 0 \][/tex]
- Case 2: [tex]\(n = 9\)[/tex]
- Here, we need [tex]\(pq = 9\)[/tex] and [tex]\(p + q = 6\)[/tex].
- The pairs that satisfy these equations are [tex]\(p = 3\)[/tex] and [tex]\(q = 3\)[/tex].
- Thus, the factored form is:
[tex]\[ (x + 3)(x + 3) = 0 \quad \text{or} \quad (x + 3)^2 = 0 \][/tex]
- Case 3: [tex]\(n = 7\)[/tex]
- Now, we need [tex]\(pq = 7\)[/tex] and [tex]\(p + q = 6\)[/tex].
- Unfortunately, there are no integer pairs for [tex]\(p\)[/tex] and [tex]\(q\)[/tex] that satisfy these equations.
5. Factoring in general:
Often [tex]\(n\)[/tex] is known or can typically be simplified further. While in our cases [tex]\(n\)[/tex] was straightforward enough for direct factoring.
In conclusion, solving [tex]\(x^2 + 6x + n = 0\)[/tex] for different [tex]\(n\)[/tex] requires identifying pairs [tex]\((p, q)\)[/tex] that add up to the middle coefficient (in this case, 6) and multiply to the constant term [tex]\(n\)[/tex]. Depending on [tex]\(n\)[/tex], if integer solutions exist, factoring can be manually achieved as shown.
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