Question

problem 5, find all the zeros given function and one zero of the function. f(x)=x^3 + 5x^2 - 77x - 441 with knowledge that x = - 7 is a zero of the polynomial function.

Explanation:

Step1: Use synthetic division

Since \(x = - 7\) is a zero, divide \(f(x)=x^{3}+5x^{2}-77x - 441\) by \(x + 7\) using synthetic - division.
The coefficients are \(1,5,-77,-441\).
The synthetic - division setup:

-7 |  1   5   -77  -441
    |     -7    14   441
    |________________
      1  -2   -63    0

The quotient is \(x^{2}-2x - 63\).

Step2: Factor the quotient

Factor the quadratic \(x^{2}-2x - 63\).
We need to find two numbers that multiply to \(-63\) and add up to \(-2\). The numbers are \(-9\) and \(7\).
So \(x^{2}-2x - 63=(x - 9)(x + 7)\).

Step3: Find the zeros

Set \(f(x)=(x + 7)(x^{2}-2x - 63)=(x + 7)(x - 9)(x+7)=0\).
Using the zero - product property \(a\cdot b\cdot c = 0\) implies \(a = 0\) or \(b = 0\) or \(c = 0\).
\(x+7 = 0\) gives \(x=-7\), and \(x - 9=0\) gives \(x = 9\).

Answer:

The zeros of the function \(f(x)=x^{3}+5x^{2}-77x - 441\) are \(x=-7\) and \(x = 9\).