Solve the following system 2x^3 - 6x^2 - 6x +18 = 0.
a. factor and solve
b. how many solution does this equation have?
c. Is there any indication in the function that might indicate how many solutions
d. Were there any constants factored out in the steps done in part a? What was the impact of factoring out that constant on the value or quantity of the solutions?
I got x equals the square root of 3 and 3, is that right? and please help for the other parts, thank you very much!
a. factor and solve
b. how many solution does this equation have?
c. Is there any indication in the function that might indicate how many solutions
d. Were there any constants factored out in the steps done in part a? What was the impact of factoring out that constant on the value or quantity of the solutions?
I got x equals the square root of 3 and 3, is that right? and please help for the other parts, thank you very much!
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a.) 2x³ - 6x² - 6x + 18 = 0
2(x - 3)(x² - 3) = 0
If the product of three factors equals zero, then any number of factors equals zero.
2 ≠ 0
If x - 3 = 0,
x = 3
If x² - 3 = 0.
x² = 3
x = ± √3
Solutions { - √3, √3, 3 }
b.) There are three solutions.
c.) The value of the largest exponent in the function generally gives an indication of the number of solutions.
c.) The factor 2 is a constant, so it does not effect the solution.
2(x - 3)(x² - 3) = 0
If the product of three factors equals zero, then any number of factors equals zero.
2 ≠ 0
If x - 3 = 0,
x = 3
If x² - 3 = 0.
x² = 3
x = ± √3
Solutions { - √3, √3, 3 }
b.) There are three solutions.
c.) The value of the largest exponent in the function generally gives an indication of the number of solutions.
c.) The factor 2 is a constant, so it does not effect the solution.
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Rewrite as:
x^3-3x^2-3x+9=0
Check if the multiples of 9 or 1 are solutions. You got 3, which is right. Factor out (r-3) using long division or something, you get: x^2-3=0 You can easily solve for this and you get x=+/-sqrt(3). So the cubic can be written as (x-3)(x+sqrt(3))(x-sqrt(3)). There is a theorem that says a polynomial of n degrees has n solutions. We are talking real coefficients here, so you know this cubic will have 3 solutions. It always has at least one real solution since complex solutions come in conjugate pairs. Since you figured out 3 and sqrt(3), you know there is one more real solution, since there can't be an odd number of complex roots and a cubic has 3 roots. And for d) yeah you factored out the multiplicative 2, which makes no difference. Sorry for incoherent presentation.
x^3-3x^2-3x+9=0
Check if the multiples of 9 or 1 are solutions. You got 3, which is right. Factor out (r-3) using long division or something, you get: x^2-3=0 You can easily solve for this and you get x=+/-sqrt(3). So the cubic can be written as (x-3)(x+sqrt(3))(x-sqrt(3)). There is a theorem that says a polynomial of n degrees has n solutions. We are talking real coefficients here, so you know this cubic will have 3 solutions. It always has at least one real solution since complex solutions come in conjugate pairs. Since you figured out 3 and sqrt(3), you know there is one more real solution, since there can't be an odd number of complex roots and a cubic has 3 roots. And for d) yeah you factored out the multiplicative 2, which makes no difference. Sorry for incoherent presentation.