Answer:
x = 1, x = 2, x = 5
Step-by-step explanation:
[tex]f(x)=x^3-8x^2+17x-10[/tex]
if [tex]x - 2[/tex] is a factor of [tex]f(x)[/tex], then
[tex]f(x)=(x-2)(x^2+bx+5)[/tex]
Expand brackets:
[tex]\implies f(x)=x^3+bx^2+5x-2x^2-2bx-10[/tex]
Combine like terms:
[tex]\implies f(x)=x^3+(b-2)x^2+(5-2b)x-10[/tex]
Compare coefficients for [tex]x^2[/tex]:
[tex]b-2=-8[/tex]
[tex]\implies b=-6[/tex]
Therefore,
[tex]f(x)=(x-2)(x^2-6x+5)[/tex]
Factoring [tex](x^2-6x+5)[/tex]:
[tex]\implies f(x)=(x-2)(x-1)(x-5)[/tex]
To find the zeros of [tex]f(x)[/tex], set the function to zero and solve for [tex]x[/tex]:
[tex]\implies f(x)=0\\\\\implies(x-2)(x-1)(x-5)=0\\\\\implies x=2,x=1,x=5[/tex]
