Question

How do you simplify $\sqrt {0.25{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, $0.25$. Then, we write down all the variables out, instead of using exponents. We will get the completely factored monomial of $0.25{x^6}$. 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 {0.25{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 $0.25{x^6}$.
First step is to take prime factorization of the coefficient, $0.25$. We can write this as $\dfrac{{25}}{{100}}$.
$0.25 = \dfrac{{25}}{{100}}$
Then, we can break $25$ into $5 \times 5$, and break $100$ into $2 \times 2 \times 5 \times 5$.
$0.25 = \dfrac{{5 \times 5}}{{2 \times 2 \times 5 \times 5}}$
So, the prime factorization of $0.25$ is:
$0.25 = \dfrac{{5 \times 5}}{{2 \times 2 \times 5 \times 5}}$
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 $0.25{x^6}$ is:
$0.25{x^6} = \dfrac{{5 \times 5}}{{2 \times 2 \times 5 \times 5}} \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$
Now, pair the prime factors and take the product of prime factors, choosing one factor out of every pair.
\[0.25{x^6} = \dfrac{{\underline {5 \times 5} }}{{\underline {2 \times 2} \times \underline {5 \times 5} }} \cdot \underline {x \cdot x} \cdot \underline {x \cdot x} \cdot \underline {x \cdot x} \]
\[ \Rightarrow \sqrt {0.25{x^6}} = \dfrac{5}{{10}} \cdot x \cdot x \cdot x\]
$\therefore \sqrt {0.25{x^6}} = 0.5{x^3}$

Hence, $\sqrt {0.25{x^6}} = 0.5{x^3}$.

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: $0.25{x^6}$
Completely factored monomial:
$\dfrac{{5 \times 5}}{{2 \times 2 \times 5 \times 5}} \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$
Factored monomial:
$\dfrac{{5 \times 5}}{{2 \times 5 \times 10}} \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$
Notice that each part of the completely factored monomial, $\dfrac{{5 \times 5}}{{2 \times 2 \times 5 \times 5}} \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$, is completely factored. For example, we can’t factor $2,5$ or $x$ any more than it already is.
Also, we can notice that each part of the factored monomial, $\dfrac{{5 \times 5}}{{2 \times 5 \times 10}} \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$, is not necessarily completely factored. For example, $x$ is completely factored, but $10$ is not.