Pergunta

26. Which binomial IS afactor of x^3+6x^2+9x+54 (x-6) (x-3) (x^2+9) (x+3)
Solução
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JoaquimVeterano · Tutor por 11 anos
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To determine which binomial is a factor of the polynomial x^3 + 6x^2 + 9x + 54, we can use the Factor Theorem. The Factor Theorem states that if x = a is a root of the polynomial, then (x - a) is a factor of the polynomial.
Let's test each binomial to see if it is a factor:
1. **Test (x - 6):**
Substitute x = 6 into the polynomial:
6^3 + 6 \cdot 6^2 + 9 \cdot 6 + 54 = 216 + 216 + 54 + 54 = 540 \neq 0
So, (x - 6) is not a factor.
2. **Test (x - 3):**
Substitute x = 3 into the polynomial:
3^3 + 6 \cdot 3^2 + 9 \cdot 3 + 54 = 27 + 54 + 27 + 54 = 162 \neq 0
So, (x - 3) is not a factor.
3. **Test (x^2 + 9):**
Substitute x = 0 into the polynomial:
0^3 + 6 \cdot 0^2 + 9 \cdot 0 + 54 = 54 \neq 0
So, (x^2 + 9) is not a factor.
4. **Test (x + 3):**
Substitute x = -3 into the polynomial:
(-3)^3 + 6 \cdot (-3)^2 + 9 \cdot (-3) + 54 = -27 + 54 - 27 + 54 = 54 \neq 0
So, (x + 3) is not a factor.
Since none of the substitutions resulted in zero, it appears there was an error in the initial polynomial factorization. Let's re-evaluate the polynomial by factoring it directly.
We can try factoring by grouping:
x^3 + 6x^2 + 9x + 54
Group the terms:
(x^3 + 6x^2) + (9x + 54)
Factor out the common terms:
x^2(x + 6) + 9(x + 6)
Factor out the common binomial factor (x + 6):
(x^2 + 9)(x + 6)
So, the factors of the polynomial are (x^2 + 9) and (x + 6).
Therefore, the correct answer is:
(x^2 + 9)
Let's test each binomial to see if it is a factor:
1. **Test (x - 6):**
Substitute x = 6 into the polynomial:
6^3 + 6 \cdot 6^2 + 9 \cdot 6 + 54 = 216 + 216 + 54 + 54 = 540 \neq 0
So, (x - 6) is not a factor.
2. **Test (x - 3):**
Substitute x = 3 into the polynomial:
3^3 + 6 \cdot 3^2 + 9 \cdot 3 + 54 = 27 + 54 + 27 + 54 = 162 \neq 0
So, (x - 3) is not a factor.
3. **Test (x^2 + 9):**
Substitute x = 0 into the polynomial:
0^3 + 6 \cdot 0^2 + 9 \cdot 0 + 54 = 54 \neq 0
So, (x^2 + 9) is not a factor.
4. **Test (x + 3):**
Substitute x = -3 into the polynomial:
(-3)^3 + 6 \cdot (-3)^2 + 9 \cdot (-3) + 54 = -27 + 54 - 27 + 54 = 54 \neq 0
So, (x + 3) is not a factor.
Since none of the substitutions resulted in zero, it appears there was an error in the initial polynomial factorization. Let's re-evaluate the polynomial by factoring it directly.
We can try factoring by grouping:
x^3 + 6x^2 + 9x + 54
Group the terms:
(x^3 + 6x^2) + (9x + 54)
Factor out the common terms:
x^2(x + 6) + 9(x + 6)
Factor out the common binomial factor (x + 6):
(x^2 + 9)(x + 6)
So, the factors of the polynomial are (x^2 + 9) and (x + 6).
Therefore, the correct answer is:
(x^2 + 9)
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