What are the solutions to the equation 3(2-x)^3-2(2-x)^2-19(2-x)-6=0 ?

The solutions to the equation 3(2-x)^3-2(2-x)^2-19(2-x)-6=0 are x=-1,x= 7/3 ,x=4

Find the zeros of 3(2-x)^3-2(2-x)^2-19(2-x)-6=0

The zeros of 3(2-x)^3-2(2-x)^2-19(2-x)-6=0 are x=-1,x= 7/3 ,x=4

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3 (2 - x)^{3} - 2 (2 - x)^{2} - 19 (2 - x) - 6 = 0

x = - 1, x = \frac{7}{3}, x = 4

Decimal Notation

x = - 1, x = 2.33333 \dots, x = 4

Solution steps

3 (2 - x)^{3} - 2 (2 - x)^{2} - 19 (2 - x) - 6 = 0

Expand

3 (2 - x)^{3} - 2 (2 - x)^{2} - 19 (2 - x) - 6 : - 3 x^{3} + 16 x^{2} - 9 x - 28

- 3 x^{3} + 16 x^{2} - 9 x - 28 = 0

Factor

- 3 x^{3} + 16 x^{2} - 9 x - 28 : - (x + 1) (3 x - 7) (x - 4)

- (x + 1) (3 x - 7) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or 3 x - 7 = 0 or x - 4 = 0

Solve

x + 1 = 0 : x = - 1

Solve

3 x - 7 = 0 : x = \frac{7}{3}

Solve

x - 4 = 0 : x = 4

The solutions are

x = - 1, x = \frac{7}{3}, x = 4

x^{3} + 2 x^{2} + 16 x + 32 = 0

m^{3} = \frac{125}{64}

2 x^{4} + 3 x^{3} + 16 x + 24 = 0

x^{4} - (x - 0.005)^{4} - 0. 1^{4} = 0

x^{3} + 3 x^{2} + 9 x + 27

What are the solutions to the equation 3(2-x)^3-2(2-x)^2-19(2-x)-6=0 ?
The solutions to the equation 3(2-x)^3-2(2-x)^2-19(2-x)-6=0 are x=-1,x= 7/3 ,x=4
Find the zeros of 3(2-x)^3-2(2-x)^2-19(2-x)-6=0
The zeros of 3(2-x)^3-2(2-x)^2-19(2-x)-6=0 are x=-1,x= 7/3 ,x=4