Question

this question has several parts that must be completed sequentially. if you skip a skipped part.
tutorial exercise
use the given zero to find all the zeros of the function. (enter your answers as a comma - separated list. include the given zero in your answer.)
function: ( f(x)=x^{3}+x^{2}+49x + 49 ), zero: ( 7i )
step 1
since the polynomial has real coefficients, the complex conjugate is also a zero of the function. therefore, since ( 7i ) is a zero, ( - 7i ) is also a zero of the function.
step 2
use synthetic division to divide the polynomial by the two known zeros, starting with ( 7i ).
( 7i)
skip (you cannot come back)

Explanation:

Step1: Identify conjugate zero

Since \(f(x)\) has real coefficients, if \(7i\) is a zero, then \(-7i\) is also a zero.

Step2: Set up synthetic division for \(7i\)

Write coefficients of \(f(x)=x^3+x^2+49x+49\): \(1, 1, 49, 49\)
Perform synthetic division with \(7i\):

$$\begin{array}{r|rrrr} 7i & 1 & 1 & 49 & 49 \\ & & 7i & 7i(1+7i) & 7i(49+7i+7i) \\ \hline & 1 & 1+7i & 49+7i+49i^2 & 0 \end{array}$$

Simplify \(49i^2=-49\), so \(49+7i+49i^2=7i\)

$$\begin{array}{r|rrrr} 7i & 1 & 1 & 49 & 49 \\ & & 7i & 7i-49 & 0 \\ \hline & 1 & 1+7i & 7i & 0 \end{array}$$

Step3: Divide quotient by \(-7i\)

Take quotient \(x^2+(1+7i)x+7i\), use synthetic division with \(-7i\):

$$\begin{array}{r|rrr} -7i & 1 & 1+7i & 7i \\ & & -7i & -7i(1+7i-7i) \\ \hline & 1 & 1 & 0 \end{array}$$

Simplify:

$$\begin{array}{r|rrr} -7i & 1 & 1+7i & 7i \\ & & -7i & -7i \\ \hline & 1 & 1 & 0 \end{array}$$

Step4: Find remaining zero

The resulting polynomial is \(x+1\), set to 0: \(x+1=0 \implies x=-1\)

Answer:

\(-1, -7i, 7i\)