IDNLearn.com provides a reliable platform for finding accurate and timely answers. Explore a wide array of topics and find reliable answers from our experienced community members.
Sagot :
To find the solutions to the equation [tex]\(x^3 + 6x^2 - 40x = 192\)[/tex], we need to understand where the graph of the function [tex]\( y = x^3 + 6x^2 - 40x \)[/tex] intersects the horizontal line [tex]\( y = 192 \)[/tex].
Let's rephrase the problem into one equation by setting the two equations equal to each other:
[tex]\[ x^3 + 6x^2 - 40x = 192 \][/tex]
But for clarity of understanding, we'll convert it to:
[tex]\[ x^3 + 6x^2 - 40x - 192 = 0 \][/tex]
Now, let's solve this polynomial equation step-by-step to find the values of [tex]\( x \)[/tex].
### Step-by-Step Solution
1. Equation Setup:
[tex]\[ x^3 + 6x^2 - 40x - 192 = 0 \][/tex]
2. Finding Rational Roots:
By the Rational Root Theorem, potential rational roots of the polynomial are the factors of the constant term (-192) divided by the factors of the leading coefficient (1). Hence, the possible rational roots are:
[tex]\[ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 16, \pm 24, \pm 32, \pm 48, \pm 64, \pm 96, \pm 192 \][/tex]
3. Testing Potential Roots:
Start by checking some of the potential roots using substitution.
- Testing [tex]\( x = -8 \)[/tex]:
[tex]\[ (-8)^3 + 6(-8)^2 - 40(-8) - 192 = -512 + 384 + 320 - 192 \][/tex]
[tex]\[ = 0 \][/tex]
Since this yields 0, [tex]\( x = -8 \)[/tex] is a root.
4. Polynomial Division:
Use polynomial long division or synthetic division to divide the original polynomial by [tex]\( (x + 8) \)[/tex], since [tex]\( x + 8 \)[/tex] is a factor.
Dividing [tex]\( x^3 + 6x^2 - 40x - 192 \)[/tex] by [tex]\( x + 8 \)[/tex] gives:
[tex]\[ x^3 + 6x^2 - 40x - 192 = (x + 8)(x^2 - 2x - 24) \][/tex]
5. Factoring the Quadratic Expression:
We now need to solve the quadratic equation [tex]\( x^2 - 2x - 24 = 0 \)[/tex].
To factor this quadratic, find two numbers that multiply to -24 and add to -2.
[tex]\[ x^2 - 2x - 24 = (x - 6)(x + 4) \][/tex]
6. Finding Additional Roots:
Set the factors equal to zero:
[tex]\[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \][/tex]
[tex]\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \][/tex]
### Summary and Verification
We have found the roots of the polynomial: [tex]\(x = -8\)[/tex], [tex]\(x = -4\)[/tex], and [tex]\(x = 6\)[/tex]. Verify these solutions by substituting back into the original equation:
[tex]\[ x^3 + 6x^2 - 40x - 192 = 0 \][/tex]
- For [tex]\( x = -8 \)[/tex]:
[tex]\[ (-8)^3 + 6(-8)^2 - 40(-8) - 192 = -512 + 384 + 320 - 192 = 0 \][/tex]
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ (-4)^3 + 6(-4)^2 - 40(-4) - 192 = -64 + 96 + 160 - 192 = 0 \][/tex]
- For [tex]\( x = 6 \)[/tex]:
[tex]\[ 6^3 + 6(6^2) - 40(6) - 192 = 216 + 216 - 240 - 192 = 0 \][/tex]
Thus, the solutions to the equation [tex]\( x^3 + 6x^2 - 40x = 192 \)[/tex] are:
[tex]\[ \boxed{x = -8, x = -4, x = 6} \][/tex]
Let's rephrase the problem into one equation by setting the two equations equal to each other:
[tex]\[ x^3 + 6x^2 - 40x = 192 \][/tex]
But for clarity of understanding, we'll convert it to:
[tex]\[ x^3 + 6x^2 - 40x - 192 = 0 \][/tex]
Now, let's solve this polynomial equation step-by-step to find the values of [tex]\( x \)[/tex].
### Step-by-Step Solution
1. Equation Setup:
[tex]\[ x^3 + 6x^2 - 40x - 192 = 0 \][/tex]
2. Finding Rational Roots:
By the Rational Root Theorem, potential rational roots of the polynomial are the factors of the constant term (-192) divided by the factors of the leading coefficient (1). Hence, the possible rational roots are:
[tex]\[ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 16, \pm 24, \pm 32, \pm 48, \pm 64, \pm 96, \pm 192 \][/tex]
3. Testing Potential Roots:
Start by checking some of the potential roots using substitution.
- Testing [tex]\( x = -8 \)[/tex]:
[tex]\[ (-8)^3 + 6(-8)^2 - 40(-8) - 192 = -512 + 384 + 320 - 192 \][/tex]
[tex]\[ = 0 \][/tex]
Since this yields 0, [tex]\( x = -8 \)[/tex] is a root.
4. Polynomial Division:
Use polynomial long division or synthetic division to divide the original polynomial by [tex]\( (x + 8) \)[/tex], since [tex]\( x + 8 \)[/tex] is a factor.
Dividing [tex]\( x^3 + 6x^2 - 40x - 192 \)[/tex] by [tex]\( x + 8 \)[/tex] gives:
[tex]\[ x^3 + 6x^2 - 40x - 192 = (x + 8)(x^2 - 2x - 24) \][/tex]
5. Factoring the Quadratic Expression:
We now need to solve the quadratic equation [tex]\( x^2 - 2x - 24 = 0 \)[/tex].
To factor this quadratic, find two numbers that multiply to -24 and add to -2.
[tex]\[ x^2 - 2x - 24 = (x - 6)(x + 4) \][/tex]
6. Finding Additional Roots:
Set the factors equal to zero:
[tex]\[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \][/tex]
[tex]\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \][/tex]
### Summary and Verification
We have found the roots of the polynomial: [tex]\(x = -8\)[/tex], [tex]\(x = -4\)[/tex], and [tex]\(x = 6\)[/tex]. Verify these solutions by substituting back into the original equation:
[tex]\[ x^3 + 6x^2 - 40x - 192 = 0 \][/tex]
- For [tex]\( x = -8 \)[/tex]:
[tex]\[ (-8)^3 + 6(-8)^2 - 40(-8) - 192 = -512 + 384 + 320 - 192 = 0 \][/tex]
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ (-4)^3 + 6(-4)^2 - 40(-4) - 192 = -64 + 96 + 160 - 192 = 0 \][/tex]
- For [tex]\( x = 6 \)[/tex]:
[tex]\[ 6^3 + 6(6^2) - 40(6) - 192 = 216 + 216 - 240 - 192 = 0 \][/tex]
Thus, the solutions to the equation [tex]\( x^3 + 6x^2 - 40x = 192 \)[/tex] are:
[tex]\[ \boxed{x = -8, x = -4, x = 6} \][/tex]
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Find clear and concise answers at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.
