Question

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

Answer 1

`\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`