What are the solutions to the equation 36x^3+45x^2-7x-4=0 ?

The solutions to the equation 36x^3+45x^2-7x-4=0 are x= 1/3 ,x=-1/4 ,x=-4/3

Find the zeros of 36x^3+45x^2-7x-4=0

The zeros of 36x^3+45x^2-7x-4=0 are x= 1/3 ,x=-1/4 ,x=-4/3

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36 x^{3} + 45 x^{2} - 7 x - 4 = 0

x = \frac{1}{3}, x = - \frac{1}{4}, x = - \frac{4}{3}

Decimal Notation

x = 0.33333 \dots, x = - 0.25, x = - 1.33333 \dots

Solution steps

36 x^{3} + 45 x^{2} - 7 x - 4 = 0

Factor

36 x^{3} + 45 x^{2} - 7 x - 4 : (3 x - 1) (4 x + 1) (3 x + 4)

(3 x - 1) (4 x + 1) (3 x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x - 1 = 0 or 4 x + 1 = 0 or 3 x + 4 = 0

Solve

3 x - 1 = 0 : x = \frac{1}{3}

Solve

4 x + 1 = 0 : x = - \frac{1}{4}

Solve

3 x + 4 = 0 : x = - \frac{4}{3}

The solutions are

x = \frac{1}{3}, x = - \frac{1}{4}, x = - \frac{4}{3}

x^{5} + 3 x - 4 = 0

z^{6} - z^{3} - 2 = 0

x^{3} - 9 x^{2} - 25 x + 33 = 0

3 x^{3} - 5 x^{2} + 2 = 0

x (x^{2} + 4) = 0

What are the solutions to the equation 36x^3+45x^2-7x-4=0 ?
The solutions to the equation 36x^3+45x^2-7x-4=0 are x= 1/3 ,x=-1/4 ,x=-4/3
Find the zeros of 36x^3+45x^2-7x-4=0
The zeros of 36x^3+45x^2-7x-4=0 are x= 1/3 ,x=-1/4 ,x=-4/3