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

1 Answer
Oct 4, 2017

Answer:

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) #
[Ans]