Is $$ x-6 $$ a factor of $$ 3 x^{3}-16 x^{2}-72 $$ ?

Question

UpStudy AI · Accepted Answer

To determine whether $$ x - 6 $$ is a factor of the polynomial $$ 3x^3 - 16x^2 - 72 $$, we can use the **Factor Theorem**. The Factor Theorem states that $$ x - a $$ is a factor of a polynomial $$ P(x) $$ if and only if $$ P(a) = 0 $$.

Let's apply this theorem step by step:

1. **Identify $$ a $$:**
   Here, $$ a = 6 $$ because we are checking if $$ x - 6 $$ is a factor.

2. **Evaluate $$ P(6) $$:**
   Substitute $$ x = 6 $$ into the polynomial:
   $$
   P(6) = 3(6)^3 - 16(6)^2 - 72
   $$
   
3. **Calculate each term:**
   - $$ 6^3 = 216 $$
   - $$ 6^2 = 36 $$
   - Plugging these into the equation:
     $$
     P(6) = 3 \times 216 - 16 \times 36 - 72
     $$
     $$
     P(6) = 648 - 576 - 72
     $$
     $$
     P(6) = 648 - 648 = 0
     $$
   
4. **Conclusion:**
   Since $$ P(6) = 0 $$, by the Factor Theorem, $$ x - 6 $$ is indeed a factor of the polynomial $$ 3x^3 - 16x^2 - 72 $$.

**Final Answer:**  
Yes, $$ x - 6 $$ is a factor because substituting $$ x = 6 $$ into the polynomial yields zero.
$$ x - 6 $$ is a factor of $$ 3x^3 - 16x^2 - 72 $$ because substituting $$ x = 6 $$ into the polynomial results in zero.

Is a factor of ?

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Algebra Questions