Question

What are all the possible rational zeros for `f(x)=4x^3+7x^2-11x-18` and how do you find all zeros?

Answer 1

Zeros are `x= -2 ,x ~~ -1.38 , x ~~ 1.63`

Explanation:

`f(x) = 4x^3+7x^2-11x-18` . Here constant number is `18` and

leading coefficient is `4` . Factors of `18` are `1,2,3,6,9,18` and

factors of `4` are `1,2,4 :.` Possible rational zeros are

`+-(1,2,3,6,9,18)/( 1,2,4 ) :.`

Possible rational zeros are `+-( 1,2,3,6,9,18,2/4,3/2,3/4,9/2,9/4)`

`f(1)=-18 , f(-1)=-4 , f(2)=20 , f(-2)=0 :.`

`(x+2)` is a factor. By synthetic division we can check it .

  -2    | (4)  ----   (7)   -----    (-11) --- (-18)

           | -----   (-8)  ------    (2)  ----   (18)

          |(4)  -----  ( -1)   -----   ( -9)  ----    (0)

So `f(x) = 4x^3+7x^2-11x-18 = (x+2) ( 4x^2-x-9)`

Let `g(x) = 4x^2-x-9 ; a= 4 , b= -1 , c= -9`

Discriminant `D =b^2-4ac= 145`

`x= (-b+- sqrtD)/(2a)= (1 +- sqrt145)/8 :.`

`x ~~ -1.38(2dp) , x ~~ 1.63(2dp)`

Hence zeros are `x= -2 ,x ~~ -1.38(2dp) , x ~~ 1.63(2dp)`
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