Question

How do you solve `4y^3-2=y-8y^2` by factoring and then using the zero-product principle?

Answer 1

`"Soln. Set="{-2,-1/2,1/2}"`.

Explanation:

The given eqn. is, `4y^3-2=y-8y^2.`

`:. 4y^3-2+8y^2-y=0.`

`:. ul(4y^3+8y^2)-ul(y-2)=0.`

`:. 4y^2(y+2)-1(y+2)=0.`

`:. (y+2)(4y^2-1)=0.`

`:. (y+2)(2y+1)(2y-1)=0.`

`:. y+2=0, or 2y+1=0, or 2y-1=0.`

`:. y=-2, or y=-1/2, or y=1/2.`

These roots satisfy the given eqn.

Hence, the `"Soln. Set="{-2,-1/2,1/2}"`.

Answer 2

`y=-2,y=+-1/2`

Explanation:

`"rearrange and equate to zero"`

`rArr4y^3+8y^2-y-2=0larrcolor(blue)"factorise by grouping"`

`color(red)(4y^2)(y+2)color(red)(-1)(y+2)=0`

`rArr(y+2)(color(red)(4y^2-1))=0`

`4y^2-1color(blue)" is a difference of squares"`

`•color(white)(x)a^2-b^2=(a-b)(a+b)`

`rArr(y+2)(2y-1)(2y+1)=0`

`"equate each factor to zero (zero-product principle) and"`
`"solve for y"`

`y+2=0rArry=-2`

`2y-1=0rArry=1/2`

`2y+1=0rArry=-1/2`

Answer 3

`y=-2 or y=1/2 or y=-1/2`

Explanation:

First, we put all the terms on one side.
`4y^3-2=y-8y^2`
`4y^3+8y^2-y-2=0`

Then, we group the factorize it by using common factor.
`4y^2(y+2)-1(y+2)=0`
`(y+2)(4y^2-1)=0`

Recall that the zero-product principle tells, when `ab=0`, then either `a=0` or `b=0`, or both `a` and `b` are `0`.

`(y+2)(4y^2-1)=0`
`y+2=0 or 4y^2-1=0`
`y=-2 or y^2=1/4`
`y=-2 or y=1/2 or y=-1/2`

Reminder: `sqrt (1/4)=+-1/2` , there will be 2 answers for square root

Here is the answer :)
Hope it can help you.