Find a value of y for which the expression (1-y)(3-y)(6-y) has the given value: -70, 120, and 3 1/8?

Question

Any help would be greatly appreciated!

Answer 1

I'm not sure if the last number is (31/8) or 3 + (1/8) or something else.  But I can show you how to solve the first two.

First, expand the expression.

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

(3 - y - 3y + y²)(6 - y) =

(3 - 4y + y²)(6 - y) =

18 - 3y - 24y + 4y² + 6y² - y³ =

-y³ + 10y² - 27y + 18

Now set this equal to -70 and solve for y.

-y³ + 10y² - 27y + 18 = -70

-y³ + 10y² - 27y + 18 + 70 = 0

-y³ + 10y² - 27y + 88 = 0  

-(y³ - 10y² + 27y - 88) = 0   ...factor the left side

-(y - 8)(y² - 2y + 11) = 0

Then y = 8 is a solution.

To check this let y = 8 then

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

(1-8)(3-8)(6-8) =

(-7)(-5)(-2) =

35*(-2) =

-70

This time set the expression equal to 120 and solve for y.

-y³ + 10y² - 27y + 18 = 120

-y³ + 10y² - 27y + 18 - 120 = 0

-y³ + 10y² - 27y  - 102 = 0

-y³ + 10y² - 27y  - 102 = 0

-(y³ - 10y² + 27y + 102) = 0   ...factor left side

-(y + 2)(y² - 12y + 51) = 0

Then y = -2 is a solution.

To check this let y = -2 then

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

(1+2)(3+2)(6+2) =

(3)(5)(8) =

15*8 =

120

Hope this helps you!