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If f(x)=x^3-12x^2+35x-24f(x)=x 3 −12x 2 +35x−24 and f(8)=0f(8)=0, then find all of the zeros of f(x)f(x) algebraically.

Sagot :

MrRoyal

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)

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