Question

Which of the following is(are) possible factorization(s) of $6x^3?$

• A. $6\cdot x^3$
• B. $6x\cdot x^2$
• C. $6\cdot x\cdot x\cdot x$
• D. $2\cdot 3\cdot x\cdot x\cdot x$

Answer 1

Answer: (A), (B), (C), (D)

$2\cdot 3\cdot x\cdot x\cdot x$

$\underset{–}{\text{Factor of a Monomial}:}$
To factor a monomial means to express it as a product of two or more monomials.

For example, below are several possible factorizations of $6x^4$.

• $6x^4=2x\cdot 3x^3$
• $6x^4=6x^2\cdot x^2$
• $6x^4=3x\cdot 2x\cdot x\cdot x$
• $6x^4=2\cdot 3\cdot x\cdot x\cdot x\cdot x$

$\underset{–}{\text{Note}:}$
Product of all the factors of a monomial is equal to the monomial itself.

Let's have a look at each choice separately now:
$\bullet$ $6\cdot x^3=6x^3$

$\bullet$ $6x\cdot x^2=6x^{(1+2)}=6x^3$ $(\because a^m\cdot a^n=a^{(m+n)})$

$\bullet$ $6\cdot x\cdot x\cdot x=6x^{(1+1+1)}=6x^3$ $(\because a^p\cdot a^q\cdot a^r=a^{(p+q+r)})$

$\bullet$ $2\cdot 3\cdot x\cdot x\cdot x=6x^{(1+1+1)}=6x^3$ $(\because a^p\cdot a^q\cdot a^r=a^{(p+q+r)})$

As we can see, product of all the factors in each given factorization is equal to $6x^3,$ all the given options are possible factorization of $6x^3.$