Okay so I factored 2x^3-18x-2x^2+18 and got 2(x-1)(x+3)(x-3) now how do I solve for x from there?
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You cannot solve it without an equation.
You can find the zeros by 2(x-1)(x+3)(x-3)=0
x = 1, 3 and -3
You can find the zeros by 2(x-1)(x+3)(x-3)=0
x = 1, 3 and -3
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You have to add another condition in order to solve for x.
Usually, that condition will be that you want to find the values of x that make your function zero.
i.e. 2(x-1)(x+3)(x-3) = 0.
It is clear that this will be satisfied if x is 1, 3 or -3.
Usually, that condition will be that you want to find the values of x that make your function zero.
i.e. 2(x-1)(x+3)(x-3) = 0.
It is clear that this will be satisfied if x is 1, 3 or -3.
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2x^3 - 18x - 2x^2 + 18
= 2x^3 - 2x^2 - 18x + 18
= 2(x^3 - x^2 - 9x + 9)
= 2[x^2(x - 1) - 9(x - 1)]
= 2(x^2 - 9)(x - 1)
= 2(x + 3)(x - 3)(x - 1)
Roots:
x = -3
x = 1
x = 3
= 2x^3 - 2x^2 - 18x + 18
= 2(x^3 - x^2 - 9x + 9)
= 2[x^2(x - 1) - 9(x - 1)]
= 2(x^2 - 9)(x - 1)
= 2(x + 3)(x - 3)(x - 1)
Roots:
x = -3
x = 1
x = 3