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Sagot :
Sure! Let's tackle the problem step by step.
### Part (a): Finding the constants [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]
We start by expanding the polynomial [tex]\((x + 5)(x - 4)(x - a)\)[/tex] and then comparing the resulting polynomial with the given polynomial [tex]\( x^3 + bx^2 - 23x + c \)[/tex].
1. Expand the given polynomial:
[tex]\[ (x + 5)(x - 4)(x - a) \][/tex]
2. First, expand [tex]\((x + 5)(x - 4)\)[/tex]:
[tex]\[ (x + 5)(x - 4) = x^2 - 4x + 5x - 20 = x^2 + x - 20 \][/tex]
3. Next, expand [tex]\((x^2 + x - 20)(x - a)\)[/tex]:
[tex]\[ (x^2 + x - 20)(x - a) = x^3 - ax^2 + x^2 - 20x + a x - 20a \][/tex]
4. Combine like terms:
[tex]\[ x^3 + (1 - a)x^2 + (a - 20)x - 20a \][/tex]
5. Compare this with [tex]\(x^3 + bx^2 - 23x + c\)[/tex]:
From the comparison:
[tex]\[ x^3 + (1 - a)x^2 + (a - 20)x - 20a = x^3 + bx^2 - 23x + c \][/tex]
Equating coefficients, we get:
[tex]\[ 1 - a = b \][/tex]
[tex]\[ a - 20 = -23 \][/tex]
[tex]\[ -20a = c \][/tex]
6. Solve for [tex]\(a\)[/tex]:
[tex]\[ a - 20 = -23 \implies a = -3 \][/tex]
7. Solve for [tex]\(b\)[/tex]:
[tex]\[ 1 - a = b \implies 1 - (-3) = b \implies b = 4 \][/tex]
8. Solve for [tex]\(c\)[/tex]:
[tex]\[ -20a = c \implies -20(-3) = c \implies c = 60 \][/tex]
So, the values of the constants are:
[tex]\[ a = -3, \quad b = 4, \quad c = 60 \][/tex]
### Part (b): Solving the inequality [tex]\( x^3 + 4x^2 - 23x + 60 < 0 \)[/tex]
To solve the inequality [tex]\( x^3 + 4x^2 - 23x + 60 < 0 \)[/tex], we need to determine the roots of the cubic polynomial.
1. Find the roots:
Use the polynomial [tex]\(x^3 + 4x^2 - 23x + 60\)[/tex]. We will use factor theorem or other root-finding methods.
Using trial and error or other factorization methods, we find that the roots are:
[tex]\[ x = -5, \quad x = 3, \quad x = 4 \][/tex]
2. Factor the polynomial:
[tex]\[ x^3 + 4x^2 - 23x + 60 = (x + 5)(x - 3)(x - 4) \][/tex]
3. Determine where the factors change sign to solve the inequality [tex]\( (x + 5)(x - 3)(x - 4) < 0 \)[/tex]:
- Roots are [tex]\( x = -5, x = 3, x = 4 \)[/tex].
- Test intervals between the roots:
- For [tex]\( x < -5 \)[/tex]: All factors are negative or positive leading to [tex]\( > 0 \)[/tex].
- For [tex]\( -5 < x < 3 \)[/tex]: One factor is negative, meaning the product [tex]\( < 0 \)[/tex].
- For [tex]\( 3 < x < 4 \)[/tex]: Two factors are negative, resulting in product [tex]\( > 0 \)[/tex].
- For [tex]\( x > 4 \)[/tex]: All factors are positive, leading to [tex]\( > 0 \)[/tex].
4. Combine the intervals:
Therefore, the solution for the inequality is:
[tex]\[ -5 < x < 3 \][/tex]
### Summary:
- The values of the constants are:
[tex]\[ a = -3, \quad b = 4, \quad c = 60 \][/tex]
- The inequality [tex]\( x^3 + 4x^2 - 23x + 60 < 0 \)[/tex] is satisfied when:
[tex]\[ -5 < x < 3 \][/tex]
### Part (a): Finding the constants [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]
We start by expanding the polynomial [tex]\((x + 5)(x - 4)(x - a)\)[/tex] and then comparing the resulting polynomial with the given polynomial [tex]\( x^3 + bx^2 - 23x + c \)[/tex].
1. Expand the given polynomial:
[tex]\[ (x + 5)(x - 4)(x - a) \][/tex]
2. First, expand [tex]\((x + 5)(x - 4)\)[/tex]:
[tex]\[ (x + 5)(x - 4) = x^2 - 4x + 5x - 20 = x^2 + x - 20 \][/tex]
3. Next, expand [tex]\((x^2 + x - 20)(x - a)\)[/tex]:
[tex]\[ (x^2 + x - 20)(x - a) = x^3 - ax^2 + x^2 - 20x + a x - 20a \][/tex]
4. Combine like terms:
[tex]\[ x^3 + (1 - a)x^2 + (a - 20)x - 20a \][/tex]
5. Compare this with [tex]\(x^3 + bx^2 - 23x + c\)[/tex]:
From the comparison:
[tex]\[ x^3 + (1 - a)x^2 + (a - 20)x - 20a = x^3 + bx^2 - 23x + c \][/tex]
Equating coefficients, we get:
[tex]\[ 1 - a = b \][/tex]
[tex]\[ a - 20 = -23 \][/tex]
[tex]\[ -20a = c \][/tex]
6. Solve for [tex]\(a\)[/tex]:
[tex]\[ a - 20 = -23 \implies a = -3 \][/tex]
7. Solve for [tex]\(b\)[/tex]:
[tex]\[ 1 - a = b \implies 1 - (-3) = b \implies b = 4 \][/tex]
8. Solve for [tex]\(c\)[/tex]:
[tex]\[ -20a = c \implies -20(-3) = c \implies c = 60 \][/tex]
So, the values of the constants are:
[tex]\[ a = -3, \quad b = 4, \quad c = 60 \][/tex]
### Part (b): Solving the inequality [tex]\( x^3 + 4x^2 - 23x + 60 < 0 \)[/tex]
To solve the inequality [tex]\( x^3 + 4x^2 - 23x + 60 < 0 \)[/tex], we need to determine the roots of the cubic polynomial.
1. Find the roots:
Use the polynomial [tex]\(x^3 + 4x^2 - 23x + 60\)[/tex]. We will use factor theorem or other root-finding methods.
Using trial and error or other factorization methods, we find that the roots are:
[tex]\[ x = -5, \quad x = 3, \quad x = 4 \][/tex]
2. Factor the polynomial:
[tex]\[ x^3 + 4x^2 - 23x + 60 = (x + 5)(x - 3)(x - 4) \][/tex]
3. Determine where the factors change sign to solve the inequality [tex]\( (x + 5)(x - 3)(x - 4) < 0 \)[/tex]:
- Roots are [tex]\( x = -5, x = 3, x = 4 \)[/tex].
- Test intervals between the roots:
- For [tex]\( x < -5 \)[/tex]: All factors are negative or positive leading to [tex]\( > 0 \)[/tex].
- For [tex]\( -5 < x < 3 \)[/tex]: One factor is negative, meaning the product [tex]\( < 0 \)[/tex].
- For [tex]\( 3 < x < 4 \)[/tex]: Two factors are negative, resulting in product [tex]\( > 0 \)[/tex].
- For [tex]\( x > 4 \)[/tex]: All factors are positive, leading to [tex]\( > 0 \)[/tex].
4. Combine the intervals:
Therefore, the solution for the inequality is:
[tex]\[ -5 < x < 3 \][/tex]
### Summary:
- The values of the constants are:
[tex]\[ a = -3, \quad b = 4, \quad c = 60 \][/tex]
- The inequality [tex]\( x^3 + 4x^2 - 23x + 60 < 0 \)[/tex] is satisfied when:
[tex]\[ -5 < x < 3 \][/tex]
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