Question

Total number of factors when $15x^2$ is completely factored is .

Answer 1

$\underset{–}{\text{Complete Factor of a Monomial}:}$
To factor a monomial completely, we write the coefficient as a product of primes and expand the variable part.

For example, to completely factor $15x^3,$ we can write the prime factorization of $15$ as $3\cdot 5$ and write $x^3$ as $x\cdot x\cdot x$. Therefore, the complete factorization of $15x^3$:$15x^3=3\cdot 5\cdot x\cdot x\cdot x$

$\underset{–}{\text{Need to Completely Factorize:}}15x^2$


Here, Coefficient: $15$ and
Variable part: $x^2$

Prime factorization of coefficient $15:$
$9=3\cdot 5$

Expension of variable part $x^2:$
$x^2=x\cdot x$

Therefore, complete factorization of $15x^2:$ $15x^2=3\cdot 5\cdot x\cdot x$

As a result, when $15x^2$ is completely factorized, the factors are, $3,5,x$ and $x.$ Therefore, total number of factors, when $15x^2$ is completely factorized is $\underset{–}{4}.$