Pergunta

(1) 5. Which binomial is a factor of x^2+8x+8 (x+2) (x+1) Prime (x+4)
Solução
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ElisaVeterano · Tutor por 10 anos
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To determine which binomial is a factor of x^2 + 8x + 8, we need to factor the quadratic expression.
First, we look for two numbers that multiply to the constant term (8) and add up to the coefficient of the linear term (8).
The numbers that satisfy these conditions are 2 and 4 because:
2 \times 4 = 8
2 + 4 = 6
However, we need the sum to be 8, not 6. So, we need to check if there was a mistake in the problem statement or if we need to factor it differently.
Let's try factoring by grouping:
x^2 + 8x + 8
We can rewrite the quadratic as:
x^2 + 2x + 6x + 8
Now, we group the terms:
(x^2 + 2x) + (6x + 8)
Factor out the common factors from each group:
x(x + 2) + 2(3x + 4)
Since there is no common binomial factor, it appears that the quadratic x^2 + 8x + 8 is not factorable over the integers.
Therefore, the correct answer is:
\text{Prime}
First, we look for two numbers that multiply to the constant term (8) and add up to the coefficient of the linear term (8).
The numbers that satisfy these conditions are 2 and 4 because:
2 \times 4 = 8
2 + 4 = 6
However, we need the sum to be 8, not 6. So, we need to check if there was a mistake in the problem statement or if we need to factor it differently.
Let's try factoring by grouping:
x^2 + 8x + 8
We can rewrite the quadratic as:
x^2 + 2x + 6x + 8
Now, we group the terms:
(x^2 + 2x) + (6x + 8)
Factor out the common factors from each group:
x(x + 2) + 2(3x + 4)
Since there is no common binomial factor, it appears that the quadratic x^2 + 8x + 8 is not factorable over the integers.
Therefore, the correct answer is:
\text{Prime}
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