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Solve the inequality [tex]$x^3 - 7x^2 + 10x \ \textless \ 0$[/tex].

A. [tex]$(-\infty, 0) \cup (2,5)$[/tex]
B. [tex][tex]$(-\infty, 0] \cup [2,5]$[/tex][/tex]
C. [tex]$(0,2) \cup (5, \infty)$[/tex]
D. [tex]$[0,2] \cup [5, \infty)$[/tex]
E. I don't know.

Sagot :

GinnyAnswer
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]
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