Question

f(x)=2x^3+5x^2-9x-18 using zeros of polynomial function

Answer 1

Given polynomial f(x) = 2x^3+5x^2-9x-18

By the rational root therom

since leading coefficent is 2, any ratiional zero must be a divisor of constant term 18.

p = -18,q = 2

p/q = -18/2 = -9

So the possible rational zeroes are +1,-1,+3,-3,+9,-9.

Now we test each of these possibilities.

f(x) = 2x^3+5x^2-9x-18

P(1) = 2(1)^3+5(1)^2-9(1)-18 = 2+5-9-18 is not equals to 0.

P(-1) = 2(-1)^3+5(-1)^2-9(-1)-18 = -2+5+9-18 is not equals to 0.

P(3) = 2(3)^3+5(3)^2-9(3)-18 = 54+45-27-18 is not equals to 0.

P(-3) = 2(-3)^3+5(-3)^2-9(-3)-18 = -54+45+27-18 = 0

P(9) = 2(9)^3+5(9)^2-9(9)-18 = 729+405-81-18 is not equals to 0.

P(-9) = 2(-9)^3+5(-9)^2-9(-9)-18 = -729+405+81-18 is not equls to 0.

possible rational zero of above function is x = -3.

To find other possible zeros, we use

synthetic division method.

2x^2-x-6 = 0

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

2x(x-2)+3(x-2) = 0

(2x+3)(x-2) = 0

(2x+3) = 0 and x-2 = 0

2x = -3 and x = 2

x = -3/2 and x = 2

Solution of x = -3,2,-3/2