Using the conjugate zeros theorem to find all zeros of a polynomial
Find all other zeros of , given that is a zero.
(If there is more than one zero, separate them with commas.)
(1)

Ask by Carter Ruiz. in the United States

Mar 31,2025

Question

Using the conjugate zeros theorem to find all zeros of a polynomial
Find all other zeros of , given that is a zero.
(If there is more than one zero, separate them with commas.)
(1)

Ask by Carter Ruiz. in the United States

Mar 31,2025

UpStudy AI · Accepted Answer

We are given that $$2+2i$$ is a zero of
$$
P(x) = x^3 - 8x + 32.
$$
By the conjugate zeros theorem, its complex conjugate $$2-2i$$ is also a zero.

**Step 1. Form the quadratic factor.**

The factors corresponding to $$2+2i$$ and $$2-2i$$ are:
$$
\begin{aligned}
&x - (2+2i) \quad \text{and} \quad x - (2-2i).
\end{aligned}
$$
Multiply them:
$$
\begin{aligned}
(x - (2+2i))(x - (2-2i)) &= (x-2-2i)(x-2+2i) \
&= (x-2)^2 - (2i)^2 \
&= (x-2)^2 - (-4) \
&= (x-2)^2 + 4 \
&= x^2 - 4x + 4 + 4 \
&= x^2 - 4x + 8.
\end{aligned}
$$

**Step 2. Factor the polynomial by dividing.**

Since $$P(x)$$ is cubic and we have a quadratic factor $$x^2 - 4x + 8$$, we can write:
$$
P(x) = (x^2 - 4x + 8)Q(x)
$$
where $$Q(x)$$ is a linear factor of the form $$x + c$$.

Assume:
$$
(x^2 - 4x + 8)(x + c) = x^3 - 8x + 32.
$$
Expand the left side:
$$
\begin{aligned}
(x^2 - 4x + 8)(x + c) &= x^3 + cx^2 - 4x^2 - 4cx + 8x + 8c \
&= x^3 + (c-4)x^2 + (8-4c)x + 8c.
\end{aligned}
$$
Match coefficients with $$x^3 - 8x + 32$$:
- Coefficient of $$x^2$$:
  $$
  c - 4 = 0 \quad \Rightarrow \quad c = 4.
  $$
- Coefficient of $$x$$:
  $$
  8 - 4(4) = 8 - 16 = -8,
  $$
  which matches.
- Constant term:
  $$
  8c = 8(4) = 32,
  $$
  which also matches.

Thus, the linear factor is:
$$
Q(x) = x + 4.
$$

**Step 3. Write all zeros.**

The zeros of $$P(x)$$ are the solutions of:
$$
x^2 - 4x + 8 = 0 \quad \text{and} \quad x + 4 = 0.
$$

We already have the zeros of the quadratic factor:
$$
x = 2+2i \quad \text{and} \quad x = 2-2i.
$$
The third zero is given by:
$$
x + 4 = 0 \quad \Rightarrow \quad x = -4.
$$

Therefore, all zeros of the polynomial are:
$$
2+2i,\quad 2-2i,\quad -4.
$$
The zeros of the polynomial $$ P(x) = x^3 - 8x + 32 $$ are $$ 2+2i $$, $$ 2-2i $$, and $$ -4 $$.

Using the conjugate zeros theorem to find all zeros of a polynomial
Find all other zeros of , given that is a zero.
(If there is more than one zero, separate them with commas.)
(1)

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Using the conjugate zeros theorem to find all zeros of a polynomial Find all other zeros of , given that is a zero. (If there is more than one zero, separate them with commas.) (1)

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Using the conjugate zeros theorem to find all zeros of a polynomial
Find all other zeros of , given that is a zero.
(If there is more than one zero, separate them with commas.)
(1)