Determine the zeros of the curve of `y=2x^3-3x^2-12x+20 ` ``

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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`y= 2x^3 - 3x^2 - 12x + 20`

 To find the zeros, first we will try to substitute the factors of 20

Factors are: `+-1, +-2, +-4 , +-5 , +-10` , ...etc

x= 2

`==gt y== 2(2^3) - 3(2^2) - 12(2) + 20 `

`==gt y= 16 - 12 - 24 + 20 = 0`

`` Then x= 2 is one of the zeros.

==> Then (x-2) is a factor of y.

Now we will divide y by (x-2)

==>`y= (x-2)(2x^2+x-10)`

`` Now we will factor the quadratic equation.

`==gt y= (x-2)(x-2)(2x+5) `

`==gt y= (x-2)^2 (2x+5)`

`` Then we have two zeros,

`==gt x= { 2, -5/2}`

``

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