Given the function below
[tex]f(x)=x^4-11x^3+40x^2-48x^{}[/tex]
Where
[tex]x=3\text{ is a zero of the function.}[/tex]
To find the other zeros,
[tex]\begin{gathered} x=3x \\ x-3=0\text{ is a factor} \end{gathered}[/tex]
And x is common, factor out x, i.e
[tex]\begin{gathered} f(x)=x^4-11x^3+40x^2-48x^{} \\ f(x)=x(x^3-11x^2+40x-48) \end{gathered}[/tex]
Divde the function by the factor x - 3
The quotient of the function after dividing by x - 3 is
[tex]f(x)=(x-3)(x^3-8x^2+16x)[/tex]
Factor out x
[tex]\begin{gathered} f(x)=(x-3)(x^3-8x^2+16x) \\ f(x)=x(x-3)(x^2-8x+16) \end{gathered}[/tex]
Factorize the remaining equation
[tex]\begin{gathered} f(x)=x(x-3)(x^2-8x+16) \\ f(x)=x(x-3)(x^2-4x-4x+16) \\ f(x)=x(x-3)\mleft\lbrace x(x-4\mright)-4(x-4)\} \\ f(x)=x(x-3)\mleft\lbrace(x-4\mright)(x-4)\} \\ f(x)=x(x-3)(x-4)^2 \end{gathered}[/tex]
To find the zeros of the above factored function, using the zero factor principle
[tex]\begin{gathered} xy=0 \\ x=0\text{ and y}=0 \end{gathered}[/tex]
[tex]\begin{gathered} x(x-3)(x-4)^2=0 \\ x=0 \\ x-3=0 \\ x=3 \\ x-4=0 \\ x-4=0 \\ x=4 \\ x=0,3,4(\text{twice)} \end{gathered}[/tex]
Hence, the othe two zeros of the function f(x) are
[tex]x=0,x=4[/tex]
