Difficult equation...?

Question

Find all of the values of y that satisfy the equation...

(y^2+y-2)(4y-20)=0

Answer 1

To do this you have to find the values of y which make the brackets equal to zero

y^2+y-2

if you factor it

(y+2)(y-1)

with the other bracket as well that gives

y= -2,1,5

Answer 2

For the first part,

(y^2 + y - 2)

Find two numbers such that their product is -2 (Third term) and sum is +1 (middle term)

This will come out to be -2 , +1

So (y^2 - 2y + y - 2 )

Thus y (y - 2) +1 (y - 2)

(y + 1)(y - 2)

These are the factors of first part..

Second part 4y -20

Take 4 common outside the parentheses

so 4 (y - 5)

Thus would b the factors of the second part...

Answer 3

Split this into two equations:

y^2 + y - 2 = 0

4y - 20 = 0

You can factor the first equation into (y-1)(y+2), so you get three equations:

y - 1 = 0

y + 2 = 0

4y - 20 = 0 ; solving for y gives you the zeroes of the function:

y = 1

y = -2

y = 5

Answer 4

First factor (y^2+y-2): (y+2)(y-1).  So (y+2)(y-1)(4y-20)=0.

Then set each factor equal to zero.

y+2=0, y-1=0, and 4y-20=0.

Now solve each equation. You get y=-2, y=1, and y=5.

Answer 5

(y^2+y-2)=(y-2)(y+1)

so its

(y-2)(y+1)(4y-20)=0

ie. y=2 or

     y=-1 or

    y=20/4= 5

thats it

Answer 6

Can I use a calculator??

Answer 7

(y^2+y-2)(4y-20)=0 means that:

(y^2+y-2)=0   or   (4y-20)=0

> (y^2+y-2)=0 when y=1 or y=-2

> (4y-20)=0 when y=5

do you understand?

Answer 8

Close

(y+2)(y-1)(4y-20)=0

y = -2, 1 or 5

Answer 9

Factor this one step further => (y-2)(y+1)(4y-20)=0

y=2, -1 or 5