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

Zeros are $x = - 2 , x \approx - 1.38 , x \approx 1.63$

Explanation:

$f \left(x\right) = 4 {x}^{3} + 7 {x}^{2} - 11 x - 18$ . 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 $1 , 2 , 4 \therefore$ Possible rational zeros are

$\pm \frac{1 , 2 , 3 , 6 , 9 , 18}{1 , 2 , 4} \therefore$

Possible rational zeros are $\pm \left(1 , 2 , 3 , 6 , 9 , 18 , \frac{2}{4} , \frac{3}{2} , \frac{3}{4} , \frac{9}{2} , \frac{9}{4}\right)$

$f \left(1\right) = - 18 , f \left(- 1\right) = - 4 , f \left(2\right) = 20 , f \left(- 2\right) = 0 \therefore$

$\left(x + 2\right)$ is a factor. By synthetic division we can check it .

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

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

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


So $f \left(x\right) = 4 {x}^{3} + 7 {x}^{2} - 11 x - 18 = \left(x + 2\right) \left(4 {x}^{2} - x - 9\right)$

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

Discriminant $D = {b}^{2} - 4 a c = 145$

$x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{1 \pm \sqrt{145}}{8} \therefore$

$x \approx - 1.38 \left(2 \mathrm{dp}\right) , x \approx 1.63 \left(2 \mathrm{dp}\right)$

Hence zeros are $x = - 2 , x \approx - 1.38 \left(2 \mathrm{dp}\right) , x \approx 1.63 \left(2 \mathrm{dp}\right)$
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