Determine ALL of the linear factors of the polynomial $$ x^{3}-12 x^{2}+47 x-60=0 $$ when $$ x=5 $$ is a root $$ \square(x+3) $$ $$ (x-3) $$ $$ (x+5) $$ $$ \square(x-5) $$ $$ (2 x-5) $$ $$ (2 x+5) $$ $$ (x-4) $$

Question

UpStudy AI · Accepted Answer

To determine all the linear factors of the polynomial $$ P(x) = x^{3} - 12x^{2} + 47x - 60 $$ given that $$ x = 5 $$ is a root, follow these steps:

### 1. Verify $$ x = 5 $$ is a Root
Substitute $$ x = 5 $$ into $$ P(x) $$:

$$
P(5) = 5^3 - 12 \cdot 5^2 + 47 \cdot 5 - 60 = 125 - 300 + 235 - 60 = 0
$$

Since $$ P(5) = 0 $$, $$ x = 5 $$ is indeed a root. Therefore, $$ (x - 5) $$ is a linear factor.

### 2. Perform Polynomial Division
Divide $$ P(x) $$ by $$ (x - 5) $$ to find the quadratic factor.

Using **synthetic division**:

$$
\begin{array}{c|cccc}
5 & 1 & -12 & 47 & -60 \
 & & 5 & -35 & 60 \
\hline
 & 1 & -7 & 12 & 0 \
\end{array}
$$

This results in:

$$
P(x) = (x - 5)(x^2 - 7x + 12)
$$

### 3. Factor the Quadratic
Factor $$ x^2 - 7x + 12 $$:

$$
x^2 - 7x + 12 = (x - 3)(x - 4)
$$

### 4. Combine All Linear Factors
Thus, the complete factorization of $$ P(x) $$ is:

$$
P(x) = (x - 5)(x - 3)(x - 4)
$$

### **Final Linear Factors:**
- $$ (x - 5) $$
- $$ (x - 3) $$
- $$ (x - 4) $$

These are all the linear factors of the given polynomial.

**Answer:** All linear factors are (x – 5), (x – 3), and (x – 4).
All linear factors are $$ (x - 5) $$, $$ (x - 3) $$, and $$ (x - 4) $$.

Determine ALL of the linear factors of the polynomial \( x^{3}-12 x^{2}+47 x-60=0 \) when \( x=5 \) is a root \( \square(x+3) \) \( (x-3) \) \( (x+5) \) \( \square(x-5) \) \( (2 x-5) \) \( (2 x+5) \) \( (x-4) \)

Solución de tutoría real

Responder

Solución

The Deep Dive

preguntas relacionadas

Latest Algebra Questions