What are the solutions to the equation (1-y)(3-y)(6-y)=120 ?

The solutions to the equation (1-y)(3-y)(6-y)=120 are y=-2,y=6+sqrt(15)i,y=6-sqrt(15)i

Find the zeros of (1-y)(3-y)(6-y)=120

The zeros of (1-y)(3-y)(6-y)=120 are y=-2,y=6+sqrt(15)i,y=6-sqrt(15)i

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(1 - y) (3 - y) (6 - y) = 120

y = - 2, y = 6 + 15 i, y = 6 - 15 i

Solution steps

(1 - y) (3 - y) (6 - y) = 120

Expand

(1 - y) (3 - y) (6 - y) : 18 - 27 y + 10 y^{2} - y^{3}

18 - 27 y + 10 y^{2} - y^{3} = 120

Move

120

to the left side

- y^{3} + 10 y^{2} - 27 y - 102 = 0

Factor

- y^{3} + 10 y^{2} - 27 y - 102 : - (y + 2) (y^{2} - 12 y + 51)

- (y + 2) (y^{2} - 12 y + 51) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y + 2 = 0 or y^{2} - 12 y + 51 = 0

Solve

y + 2 = 0 : y = - 2

Solve

y^{2} - 12 y + 51 = 0 : y = 6 + 15 i, y = 6 - 15 i

The solutions are

y = - 2, y = 6 + 15 i, y = 6 - 15 i

3 x^{5} - 4 x^{2} - 3 = 0, 0 \leq x \leq 2

x^{3} + 5 x^{2} - 5 x - 8 = 0

x, x^{3} + y = 5

x^{3} - x^{2} - x \cdot 2^{1} = 0

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

What are the solutions to the equation (1-y)(3-y)(6-y)=120 ?
The solutions to the equation (1-y)(3-y)(6-y)=120 are y=-2,y=6+sqrt(15)i,y=6-sqrt(15)i
Find the zeros of (1-y)(3-y)(6-y)=120
The zeros of (1-y)(3-y)(6-y)=120 are y=-2,y=6+sqrt(15)i,y=6-sqrt(15)i