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Sagot :
To determine which of the given numbers are potential roots of the polynomial [tex]\( p(x) = x^3 + 6x^2 - 7x - 60 \)[/tex] according to the rational root theorem, we need to follow these steps:
1. List the Potential Roots Using the Rational Root Theorem:
The rational root theorem tells us that any rational root of the polynomial [tex]\( p(x) \)[/tex] will be a factor of the constant term divided by a factor of the leading coefficient. For the polynomial [tex]\( p(x) = x^3 + 6x^2 - 7x - 60 \)[/tex]:
- The constant term is [tex]\(-60\)[/tex].
- The leading coefficient (coefficient of [tex]\( x^3 \)[/tex]) is [tex]\( 1 \)[/tex].
Factors of [tex]\(-60\)[/tex] include: [tex]\(\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60\)[/tex].
Since the leading coefficient is [tex]\(1\)[/tex], we consider these factors as potential rational roots.
2. Evaluate Each Given Potential Root:
We now evaluate the polynomial at each of the given numbers to see which ones are actual roots (i.e., [tex]\( p(x) = 0 \)[/tex]).
- For [tex]\( x = -10 \)[/tex]:
[tex]\[ p(-10) = (-10)^3 + 6(-10)^2 - 7(-10) - 60 = -1000 + 600 + 70 - 60 = -390 \neq 0 \][/tex]
- For [tex]\( x = -7 \)[/tex]:
[tex]\[ p(-7) = (-7)^3 + 6(-7)^2 - 7(-7) - 60 = -343 + 294 + 49 - 60 = -60 \neq 0 \][/tex]
- For [tex]\( x = -5 \)[/tex]:
[tex]\[ p(-5) = (-5)^3 + 6(-5)^2 - 7(-5) - 60 = -125 + 150 + 35 - 60 = 0 \][/tex]
Hence, [tex]\( x = -5 \)[/tex] is a root of [tex]\( p(x) \)[/tex].
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ p(3) = 3^3 + 6(3)^2 - 7(3) - 60 = 27 + 54 - 21 - 60 = 0 \][/tex]
Hence, [tex]\( x = 3 \)[/tex] is a root of [tex]\( p(x) \)[/tex].
- For [tex]\( x = 15 \)[/tex]:
[tex]\[ p(15) = 15^3 + 6(15)^2 - 7(15) - 60 = 3375 + 1350 - 105 - 60 = 4560 \neq 0 \][/tex]
- For [tex]\( x = 24 \)[/tex]:
[tex]\[ p(24) = 24^3 + 6(24)^2 - 7(24) - 60 = 13824 + 3456 - 168 - 60 = 16926 \neq 0 \][/tex]
3. Conclusion:
From the evaluation above, we find that the numbers [tex]\(-5\)[/tex] and 3 are roots of the polynomial [tex]\( p(x) = x^3 + 6x^2 - 7x - 60 \)[/tex].
Thus, the numbers among the given options that are potential roots of the polynomial are:
[tex]\[ -5 \quad \text{and} \quad 3. \][/tex]
1. List the Potential Roots Using the Rational Root Theorem:
The rational root theorem tells us that any rational root of the polynomial [tex]\( p(x) \)[/tex] will be a factor of the constant term divided by a factor of the leading coefficient. For the polynomial [tex]\( p(x) = x^3 + 6x^2 - 7x - 60 \)[/tex]:
- The constant term is [tex]\(-60\)[/tex].
- The leading coefficient (coefficient of [tex]\( x^3 \)[/tex]) is [tex]\( 1 \)[/tex].
Factors of [tex]\(-60\)[/tex] include: [tex]\(\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 12, \pm 15, \pm 20, \pm 30, \pm 60\)[/tex].
Since the leading coefficient is [tex]\(1\)[/tex], we consider these factors as potential rational roots.
2. Evaluate Each Given Potential Root:
We now evaluate the polynomial at each of the given numbers to see which ones are actual roots (i.e., [tex]\( p(x) = 0 \)[/tex]).
- For [tex]\( x = -10 \)[/tex]:
[tex]\[ p(-10) = (-10)^3 + 6(-10)^2 - 7(-10) - 60 = -1000 + 600 + 70 - 60 = -390 \neq 0 \][/tex]
- For [tex]\( x = -7 \)[/tex]:
[tex]\[ p(-7) = (-7)^3 + 6(-7)^2 - 7(-7) - 60 = -343 + 294 + 49 - 60 = -60 \neq 0 \][/tex]
- For [tex]\( x = -5 \)[/tex]:
[tex]\[ p(-5) = (-5)^3 + 6(-5)^2 - 7(-5) - 60 = -125 + 150 + 35 - 60 = 0 \][/tex]
Hence, [tex]\( x = -5 \)[/tex] is a root of [tex]\( p(x) \)[/tex].
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ p(3) = 3^3 + 6(3)^2 - 7(3) - 60 = 27 + 54 - 21 - 60 = 0 \][/tex]
Hence, [tex]\( x = 3 \)[/tex] is a root of [tex]\( p(x) \)[/tex].
- For [tex]\( x = 15 \)[/tex]:
[tex]\[ p(15) = 15^3 + 6(15)^2 - 7(15) - 60 = 3375 + 1350 - 105 - 60 = 4560 \neq 0 \][/tex]
- For [tex]\( x = 24 \)[/tex]:
[tex]\[ p(24) = 24^3 + 6(24)^2 - 7(24) - 60 = 13824 + 3456 - 168 - 60 = 16926 \neq 0 \][/tex]
3. Conclusion:
From the evaluation above, we find that the numbers [tex]\(-5\)[/tex] and 3 are roots of the polynomial [tex]\( p(x) = x^3 + 6x^2 - 7x - 60 \)[/tex].
Thus, the numbers among the given options that are potential roots of the polynomial are:
[tex]\[ -5 \quad \text{and} \quad 3. \][/tex]
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