Answer:
The factors of the given function f(x) are (x-2),(x+1) and (x+3)
and zeros are 2,-1,-3
Therefore [tex]f(x)=x^3+2x^2-5x-6=(x-2)(x+1)(x+3)[/tex]
Step-by-step explanation:
Given function f is defined by [tex]f(x)=x^3+2x^2-5x-6[/tex]
To find the zeros of the given function f(x) :
First equate f(x)=0
That is [tex]f(x)=x^3+2x^2-5x-6=0[/tex]
Given that x-2 is a factor of f(x)
x-2=0
x=2
Put x=2 in [tex]f(x)=x^3+2x^2-5x-6[/tex]
[tex]f(2)=2^3+2(2)^2-5(2)-6[/tex]
[tex]=8+8-10-6[/tex]
Therefore f(2)=0 verified that x-2 is a factor
Put x=-1 in [tex]f(x)=x^3+2x^2-5x-6[/tex] we get
[tex]f(-1)=(-1)^3+2(-1)^2-5(-1)-6[/tex]
[tex]=-1+2+5-6[/tex]
[tex]=-7+7[/tex]
Therefore f(-1)=0
Therefore x+1 is a factor of f(x) and x+1=0
x=-1 is a zero of f(x)
Put x=-3 in [tex]f(x)=x^3+2x^2-5x-6[/tex] we get
[tex]f(-3)=(-3)^3+2(-3)^2-5(-3)-6[/tex]
[tex]=-27+2(9)+15-6[/tex]
[tex]=-27+18+9[/tex]
Therefore f(-3)=0
Therefore x+3 is a factor of f(x) and x+3=0
x=-3 is a zero of f(x)
Therefore the factors are (x-2),(x+1) and (x+3)
and zeros are 2,-1,-3
Therefore [tex]f(x)=x^3+2x^2-5x-6=(x-2)(x+1)(x+3)[/tex]