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3 years ago

Plz help me the question is in the picture

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Kobotan [32]3 years ago

6 0

10.5 or 10 1/2 basically ten and a half

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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.

Neporo4naja [7]

Answer:

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

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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