Question

How do you simplify $\sqrt {128{x^6}} $?

Answer 1

Hint: First, we have to write out the prime factorization of the coefficient, i.e., to take prime factorization of the coefficient, $128$. Then, we write down all the variables out, instead of using exponents. We will get the completely factored monomial of $128{x^6}$. Then, pair the prime factors and take the product of prime factors, choosing one factor out of every pair. We will get the simplified version of $\sqrt {128{x^6}} $.

Complete step by step solution:
We know that factoring is a process of breaking down a monomial into smaller terms. Although factoring a polynomial usually shortens the polynomial, factoring a polynomial expands it.
Let’s review the definition of a monomial first.
Monomial: An algebraic expression consisting of one term. Some monomials might have more than one variable.
When we factor a monomial, we are breaking it down into smaller terms. We have to express the monomial as a product of smaller monomials.
Completely factoring, or completely expanding, a monomial is a simple method. First, we determine the prime factorization of the coefficient. Then, we write down all the variables out, instead of using exponents.
Now, we look at how to factor $128{x^6}$.
First step is to take prime factorization of the coefficient, $128$. We can write this as $1 \times 128$.
$128 = 1 \times 128$
Then, we can break $128$ into $2$ and $64$.
$128 = 1 \times 2 \times 64$
Now, break $64$ into $2$ and $32$.
$128 = 1 \times 2 \times 2 \times 32$
Now, break $32$ into $2$ and $16$.
$128 = 1 \times 2 \times 2 \times 2 \times 16$
Now, break $16$ into $2$ and $8$.
$128 = 1 \times 2 \times 2 \times 2 \times 2 \times 8$
Now, break $8$ into $2$ and $4$.
$128 = 1 \times 2 \times 2 \times 2 \times 2 \times 2 \times 4$
Now, break $4$ into $2$ and $2$.
$128 = 1 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
Since, $2$ is prime, so can stop here.
So, the prime factorization of $128$ is:
$128 = 1 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
Second step is to write all the variables without exponents.
${x^6} = x \cdot x \cdot x \cdot x \cdot x \cdot x$
Finally, we can write the whole thing out. The factor of $128{x^6}$ is:
$128{x^6} = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$
Now, pair the prime factors and take the product of prime factors, choosing one factor out of every pair.
\[128{x^6} = \underline {2 \cdot 2} \cdot \underline {2 \cdot 2} \cdot \underline {2 \cdot 2} \cdot 2 \cdot \underline {x \cdot x} \cdot \underline {x \cdot x} \cdot \underline {x \cdot x} \]
$\therefore \sqrt {128{x^6}} = 8\sqrt 2 {x^3}$

Hence, $\sqrt {128{x^6}} = 8\sqrt 2 {x^3}$.

Note: Factoring a monomial, an algebraic expression consisting of one term is not the same as completely factoring it. Completely factoring a monomial is writing down its entire prime factorization of the coefficient and all the variables instead of using exponents. Factoring a monomial is more general. It means breaking down the monomial into smaller terms. These smaller terms in the monomial do not have to be completely factored themselves. For example:
Monomial: $128{x^6}$
Completely factored monomial:
$2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$
Factored monomial:
$2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4 \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$
Notice that each part of the completely factored monomial, $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$, is completely factored. For example, we can’t factor $2$ or $x$ any more than it already is.
Also, we can notice that each part of the factored monomial, $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4 \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$, is not necessarily completely factored. For example, $x$ is completely factored, but $4$ is not.