How do you find all the zeros of $f (x) = 4 x (x - 4) (x + 12) (x - \frac{1}{2})$ $f(x)=4x(x-4)(x+12)(x-1/2)$ ?

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How do you find all the zeros of $f (x) = 4 x (x - 4) (x + 12) (x - \frac{1}{2})$ $f(x)=4x(x-4)(x+12)(x-1/2)$ ?

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Answer 1 · 2016-04-18T01:58:26

$x = 0, or 4, or - 12, or \frac{1}{2}$ $x = 0, or 4, or -12, or 1/2$

Explanation:

Since $f (x)$ $f(x)$ has been factorized in the question any of it's factors set to zero will result in $f (x) = 0$ $f(x) = 0$

Hence the zeros of $f (x)$ $f(x)$ are:

Either $4 x = 0 \to x = 0$ $4x = 0 -> x = 0$
Or $x - 4 = 0 \to x = 4$ $x-4 = 0 -> x =4$
Or $x + 12 = 0 \to x = - 12$ $x+12 = 0 -> x=-12$
Or $x - \frac{1}{2} = 0 \to x = \frac{1}{2}$ $x-1/2 = 0 -> x=1/2$

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How do you find all the zeros of $f (x) = 4 x (x - 4) (x + 12) (x - \frac{1}{2})$ $f(x)=4x(x-4)(x+12)(x-1/2)$ ?

1 Answer

Explanation:

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Science

Math

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How do you find all the zeros of f(x)=4x(x−4)(x+12)(x−12)f(x)=4x(x-4)(x+12)(x-1/2)?

1 Answer

Explanation:

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How do you find all the zeros of $f (x) = 4 x (x - 4) (x + 12) (x - \frac{1}{2})$ $f(x)=4x(x-4)(x+12)(x-1/2)$ ?