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AVprozaik [17]
3 years ago
9

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.

Mathematics
1 answer:
Neporo4naja [7]3 years ago
7 0

Answer:

The zeros of f(x) are: (x - 1), (x - 3) and (x - 8)

<em></em>

Step-by-step explanation:

Given

f(x)=x^3-12x^2+35x-24

f(8) = 0

Required

Find all zeros of the f(x)

If f(8) = 0 then:

x = 8

And x - 8 is a factor

Divide f(x) by x - 8

\frac{f(x)}{x - 8} = \frac{x^3-12x^2+35x-24}{x - 8}

Expand the numerator

\frac{f(x)}{x - 8} = \frac{x^3 - 4x^2 -8x^2 + 3x + 32x - 24}{x - 8}

Rewrite as:

\frac{f(x)}{x - 8} = \frac{x^3 - 4x^2 + 3x - 8x^2 +32x - 24}{x - 8}

Factorize

\frac{f(x)}{x - 8} = \frac{(x^2 - 4x + 3)(x - 8)}{x - 8}

Expand

\frac{f(x)}{x - 8} = \frac{(x^2 -x - 3x + 3)(x - 8)}{x - 8}

Factorize

\frac{f(x)}{x - 8} = \frac{(x - 1)(x - 3)(x - 8)}{x - 8}

\frac{f(x)}{x - 8} = (x - 1)(x - 3)

Multiply both sides by x - 8

f(x) = (x - 1)(x - 3)(x - 8)

<em>Hence, the zeros of f(x) are: (x - 1), (x - 3) and (x - 8)</em>

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