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Sagot :
Let's solve the inequality [tex]\( x^3 - 7x^2 + 10x < 0 \)[/tex] step by step.
1. Factor the polynomial:
We start by factoring the cubic polynomial [tex]\( x^3 - 7x^2 + 10x \)[/tex].
[tex]\[ x^3 - 7x^2 + 10x = x(x^2 - 7x + 10) \][/tex]
2. Solve the quadratic equation inside the factor:
Next, we need to factor the quadratic [tex]\( x^2 - 7x + 10 \)[/tex]. We look for two numbers that multiply to [tex]\( 10 \)[/tex] and add up to [tex]\( 7 \)[/tex]. These numbers are [tex]\( 2 \)[/tex] and [tex]\( 5 \)[/tex].
[tex]\[ x^2 - 7x + 10 = (x - 2)(x - 5) \][/tex]
3. Combine the factors:
Thus, the cubic polynomial can be factored as:
[tex]\[ x^3 - 7x^2 + 10x = x(x - 2)(x - 5) \][/tex]
4. Determine the critical points:
The critical points (where the polynomial is zero) are obtained by setting each factor equal to zero:
[tex]\[ x = 0, \quad x = 2, \quad x = 5 \][/tex]
5. Test the intervals:
The critical points divide the number line into four intervals: [tex]\((-\infty, 0)\)[/tex], [tex]\((0, 2)\)[/tex], [tex]\((2, 5)\)[/tex], and [tex]\((5, \infty)\)[/tex]. We test a value from each interval to determine where the inequality [tex]\( x(x - 2)(x - 5) < 0 \)[/tex] holds true:
- For the interval [tex]\((-\infty, 0)\)[/tex], choose [tex]\( x = -1 \)[/tex]:
[tex]\[ (-1)(-1 - 2)(-1 - 5) = (-1)(-3)(-6) = -18 < 0 \][/tex]
- For the interval [tex]\((0, 2)\)[/tex], choose [tex]\( x = 1 \)[/tex]:
[tex]\[ (1)(1 - 2)(1 - 5) = (1)(-1)(-4) = 4 > 0 \][/tex]
- For the interval [tex]\((2, 5)\)[/tex], choose [tex]\( x = 3 \)[/tex]:
[tex]\[ (3)(3 - 2)(3 - 5) = (3)(1)(-2) = -6 < 0 \][/tex]
- For the interval [tex]\((5, \infty)\)[/tex], choose [tex]\( x = 6 \)[/tex]:
[tex]\[ (6)(6 - 2)(6 - 5) = (6)(4)(1) = 24 > 0 \][/tex]
6. Combine the intervals where the polynomial is less than zero:
From the tests, the inequality [tex]\( x(x - 2)(x - 5) < 0 \)[/tex] holds for the intervals [tex]\((-\infty, 0)\)[/tex] and [tex]\((2, 5)\)[/tex].
Therefore, the solution set of the inequality [tex]\( x^3 - 7x^2 + 10x < 0 \)[/tex] is:
[tex]\[ \boxed{(-\infty, 0) \cup (2, 5)} \][/tex]
1. Factor the polynomial:
We start by factoring the cubic polynomial [tex]\( x^3 - 7x^2 + 10x \)[/tex].
[tex]\[ x^3 - 7x^2 + 10x = x(x^2 - 7x + 10) \][/tex]
2. Solve the quadratic equation inside the factor:
Next, we need to factor the quadratic [tex]\( x^2 - 7x + 10 \)[/tex]. We look for two numbers that multiply to [tex]\( 10 \)[/tex] and add up to [tex]\( 7 \)[/tex]. These numbers are [tex]\( 2 \)[/tex] and [tex]\( 5 \)[/tex].
[tex]\[ x^2 - 7x + 10 = (x - 2)(x - 5) \][/tex]
3. Combine the factors:
Thus, the cubic polynomial can be factored as:
[tex]\[ x^3 - 7x^2 + 10x = x(x - 2)(x - 5) \][/tex]
4. Determine the critical points:
The critical points (where the polynomial is zero) are obtained by setting each factor equal to zero:
[tex]\[ x = 0, \quad x = 2, \quad x = 5 \][/tex]
5. Test the intervals:
The critical points divide the number line into four intervals: [tex]\((-\infty, 0)\)[/tex], [tex]\((0, 2)\)[/tex], [tex]\((2, 5)\)[/tex], and [tex]\((5, \infty)\)[/tex]. We test a value from each interval to determine where the inequality [tex]\( x(x - 2)(x - 5) < 0 \)[/tex] holds true:
- For the interval [tex]\((-\infty, 0)\)[/tex], choose [tex]\( x = -1 \)[/tex]:
[tex]\[ (-1)(-1 - 2)(-1 - 5) = (-1)(-3)(-6) = -18 < 0 \][/tex]
- For the interval [tex]\((0, 2)\)[/tex], choose [tex]\( x = 1 \)[/tex]:
[tex]\[ (1)(1 - 2)(1 - 5) = (1)(-1)(-4) = 4 > 0 \][/tex]
- For the interval [tex]\((2, 5)\)[/tex], choose [tex]\( x = 3 \)[/tex]:
[tex]\[ (3)(3 - 2)(3 - 5) = (3)(1)(-2) = -6 < 0 \][/tex]
- For the interval [tex]\((5, \infty)\)[/tex], choose [tex]\( x = 6 \)[/tex]:
[tex]\[ (6)(6 - 2)(6 - 5) = (6)(4)(1) = 24 > 0 \][/tex]
6. Combine the intervals where the polynomial is less than zero:
From the tests, the inequality [tex]\( x(x - 2)(x - 5) < 0 \)[/tex] holds for the intervals [tex]\((-\infty, 0)\)[/tex] and [tex]\((2, 5)\)[/tex].
Therefore, the solution set of the inequality [tex]\( x^3 - 7x^2 + 10x < 0 \)[/tex] is:
[tex]\[ \boxed{(-\infty, 0) \cup (2, 5)} \][/tex]
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