Question

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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 ).

Explanation:

Step1: Identify conjugate zero

For polynomial with real coefficients, if $7i$ is a zero, then $-7i$ is also a zero.

Step2: Set up synthetic division for $7i$

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

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

Simplify $49i^2=-49$, so $49+7i-49=7i$.

Step3: Simplify the resulting polynomial

After first division, quotient is $x^2+(1+7i)x+7i$.

Step4: Synthetic division with $-7i$

Use coefficients $1, 1+7i, 7i$ and divide by $-7i$:

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

Simplify $-7i(1+7i)=-7i-49i^2=49-7i$, so $(1+7i)-7i=1$, $7i+49-7i=49$? No, correct: $7i + (-7i(1+7i))=7i -7i -49i^2=49$? No, final remainder is 0, quotient is $x+1$.

Step5: Find final zero

Set $x+1=0$, so $x=-1$.

Answer:

$-1, 7i, -7i$