What are the solutions to the equation j(x)=3x^3,[1,1+h(x)] ?

The solutions to the equation j(x)=3x^3,[1,1+h(x)] are x=0,x=sqrt(j/3),x=-sqrt(j/3)

Find the zeros of j(x)=3x^3,[1,1+h(x)]

The zeros of j(x)=3x^3,[1,1+h(x)] are x=0,x=sqrt(j/3),x=-sqrt(j/3)

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j (x) = 3 x^{3}, [1, 1 + h (x)]

x = 0, x = \frac{j}{3}, x = - \frac{j}{3}

Solution steps

j (x) = 3 x^{3}

Move

3 x^{3}

to the left side

j x - 3 x^{3} = 0

Factor

j x - 3 x^{3} : x (j - 3 x^{2})

x (j - 3 x^{2}) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or j - 3 x^{2} = 0

Solve

j - 3 x^{2} = 0 : x = \frac{j}{3}, x = - \frac{j}{3}

The solutions are

x = 0, x = \frac{j}{3}, x = - \frac{j}{3}

6 x^{3} - 2 = 0

27 = t^{3}

(r^{2} + 1)^{4} = 0

- 5 x^{1} + 8 x^{2} + 10 x^{3} = - 8

3 x^{3} - 2 = 0

What are the solutions to the equation j(x)=3x^3,[1,1+h(x)] ?
The solutions to the equation j(x)=3x^3,[1,1+h(x)] are x=0,x=sqrt(j/3),x=-sqrt(j/3)
Find the zeros of j(x)=3x^3,[1,1+h(x)]
The zeros of j(x)=3x^3,[1,1+h(x)] are x=0,x=sqrt(j/3),x=-sqrt(j/3)