Question

The greatest common factor (GCF) of x^3, x^7, x^9 is ? Please help

Answer 1

Answer:

$x^3$

Step-by-step explanation:

Let the greatest common factor of $a,b,c$ such that $a,b,c \in\mathbb{Z}$ and they are not all equal to zero, $d$ is the common divisor of $a$ and $b$. Therefore, $d \mid a$ and $d \mid b$

You can write

$D(a) = \{d \in\mathbb{Z} : d \mid a\}$

$D(b) = \{d \in\mathbb{Z} : d \mid b\}$

The greatest common factor of $a,b$ is given as

$D(a,b) = \{d \in\mathbb{Z} : d \mid a \text{ and } d \mid b\}$

and

$D(a, b) = D(a) \cap D(b)$

This happens because $D(a,b)$ is upper bounded because if $a\neq 0$ then $d \leq |a|$

for all $d\in D(a, b)$. Therefore, the set $D(a, b)$ has the greatest elements.

Taking $x^3, x^7, x^9$ such that $x>0$

You can note that $x^7 = x^3 \cdot x^4$ and $x^9 = x^3 \cdot x^6 = x^3 \cdot x^3 \cdot x^3$

Therefore, the greatest common factor is $x^3$

Note: $x^3 \cap x^7 \cap x^9 = x^3$