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If f(x)=x^3-7x^2-14x+48 and f(8)=0, then find all of the zeros of f(x) algebraically

User Kewanna
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Answer:


\textsf{Zeros}: \quad x=8, \quad x=2, \quad x=-3

Explanation:

Given function:


f(x)=x^3-7x^2-14x+48

Factor Theorem

If f(x) is a polynomial, and f(a) = 0, then (x – a) is a factor of f(x).

Therefore, since f(8) = 0, then (x - 8) is a factor of the function f(x):


\implies f(x)=(x-8)(ax^2+bx+c)

The coefficient of x³ is a.


\implies a=1

The constant 48 is equal to -8c.


\implies c=-6

To find the value of b:


\implies cx - 8bx = -14x


\implies -6-8b=-14


\implies -8b=-8


\implies b=1

Therefore:


\implies f(x)=(x-8)(x^2+x-6)

Factor (x² + x - 6):


\implies x^2+3x-2x-6


\implies x(x+3)-2(x+3)


\implies (x-2)(x+3)

Therefore the fully factored function is:


\implies f(x)=(x-8)(x-2)(x+3)

To find the zeros of the function, set each factor to zero:


\implies x-8=0 \implies x=8


\implies x-2=0 \implies x=2


\implies x+3=0 \implies x=-3

User Fresh
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