Answer:
The zeros of f(x) are: (x - 1), (x - 3) and (x - 8)
Step-by-step explanation:
Given
[tex]f(x)=x^3-12x^2+35x-24[/tex]
[tex]f(8) = 0[/tex]
Required
Find all zeros of the f(x)
If [tex]f(8) = 0[/tex] then:
[tex]x = 8[/tex]
And [tex]x - 8[/tex] is a factor
Divide f(x) by x - 8
[tex]\frac{f(x)}{x - 8} = \frac{x^3-12x^2+35x-24}{x - 8}[/tex]
Expand the numerator
[tex]\frac{f(x)}{x - 8} = \frac{x^3 - 4x^2 -8x^2 + 3x + 32x - 24}{x - 8}[/tex]
Rewrite as:
[tex]\frac{f(x)}{x - 8} = \frac{x^3 - 4x^2 + 3x - 8x^2 +32x - 24}{x - 8}[/tex]
Factorize
[tex]\frac{f(x)}{x - 8} = \frac{(x^2 - 4x + 3)(x - 8)}{x - 8}[/tex]
Expand
[tex]\frac{f(x)}{x - 8} = \frac{(x^2 -x - 3x + 3)(x - 8)}{x - 8}[/tex]
Factorize
[tex]\frac{f(x)}{x - 8} = \frac{(x - 1)(x - 3)(x - 8)}{x - 8}[/tex]
[tex]\frac{f(x)}{x - 8} = (x - 1)(x - 3)[/tex]
Multiply both sides by x - 8
[tex]f(x) = (x - 1)(x - 3)(x - 8)[/tex]
Hence, the zeros of f(x) are: (x - 1), (x - 3) and (x - 8)
