Find the zeros of: `y=4x^3+4x^2-7x+2` ``I need to know how to do the work. PLEASE HELP!

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

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`4x^3 +4x^2 - 7x +2 = 0`

1) We will rewrite -7x = -x -6x then we will re-arrange terms.

`4x^3 + 4x^2 -x -6x +2 = 0`

`(4x^3 -x) + (4x^2 -6x+2)= 0`

`x(4x^2-1) + (2x-1)(2x-2) = 0`

`x(2x-1)(2x+1) + (2x-1)(2x-2)= 0`

`(2x-1) ( x(2x+1) + (2x-2)= 0`

`==> (2x-1)(2x^2 + x + 2x -2)= 0`

`==> (2x-1)(2x^2 +3x -2)= 0`

`==> (2x-1)(2x-1)(x+2)= 0`

`==> (2x-1)^2 (x+2)= 0`

`==> x = 1/2 `

`==> x = -2`

`==> x = (1/2, -2}`

2) We try to substitute the factor of 2 = -1, 1, -2, and 2 to find possible zeros.

`==> x = -2 ==> y(-2)= 0`

Then x = -2 is a zero of y, then, (x+2) is a factor.

`==> (4x^3+4x^2 -7x+2) = (x+2)(4x^2 -4x+1)`

`==> (4x^3 +4x^2 -7x+2)= (x+2)(2x-1)^2`

`==> x1= -2`

`==> x2= 1/2`

`==> x = {-2, 1/2}`

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