Simplify the Rational Expression: (2x-1)/(16x²-8x) Step-by-Step

Q: Why can't I just cancel the 2x from the numerator with part of the denominator?

You can only cancel complete factors, not individual terms! The expression $ 2x-1 $ is one complete factor, and $ 16x^2-8x $ must be factored completely before any canceling.

Q: Can I distribute before factoring instead?

No! For simplifying rational expressions, you want to factor, not distribute. Distributing makes expressions more complex and harder to simplify.

Q: What domain restrictions should I consider?

The denominator cannot equal zero, so $ 8x ≠ 0 $, which means $ x ≠ 0 $. Always check what values make the original denominator zero!

Rational Expression Simplification with Common Factorization

Simplify the following expression:

$\frac{2x-1}{16x^2-8x}$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Simply

00:07 Let's break down 16 into factors 2 and 8

00:10 Let's break down the exponent into multiplications

00:17 Let's mark the common factors

00:35 Let's take out the common factors from the parentheses

00:49 Let's reduce what we can

00:55 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Simplify the following expression:

$\frac{2x-1}{16x^2-8x}$

Step-by-step solution

Let's simplify the given expression:

$\frac{2x-1}{16x^2-8x}$ Remember that we can reduce complete expressions only when both the numerator and denominator are completely factored into multiplication expressions,

For this, we'll use factorization, identify that in the denominator we can factor out a common term, do this, then reduce the expressions possible in the fraction that we obtained:

$\frac{2x-1}{16x^2-8x} \\ \frac{2x-1}{8x(2x-1)} \\ \downarrow\\ \boxed{\frac{1}{8x}}$ Therefore, the correct answer is answer C.

Final Answer

$\frac{1}{8x}$

Key Points to Remember

Essential concepts to master this topic

Factorization Rule: Both numerator and denominator must be completely factored first
Common Factor Method: Factor out 8x from $16x^2-8x$ to get $8x(2x-1)$
Cancel Check: Verify $(2x-1)$ appears in both parts before canceling ✓

Common Mistakes

Avoid these frequent errors

Canceling terms instead of factors
Don't cancel 2x from numerator with 16x² in denominator = wrong simplification! This violates basic algebraic rules because you're canceling terms, not factors. Always factor completely first, then cancel only identical factors that appear in both numerator and denominator.

Practice Quiz

Test your knowledge with interactive questions

Identify the field of application of the following fraction:

$\frac{7}{13+x}$

FAQ

Everything you need to know about this question

Why can't I just cancel the 2x from the numerator with part of the denominator?

You can only cancel complete factors, not individual terms! The expression $2x-1$ is one complete factor, and $16x^2-8x$ must be factored completely before any canceling.

How do I know when an expression is completely factored?

An expression is completely factored when it's written as a product of simpler expressions that cannot be factored further. Look for common factors first, like the 8x we pulled out of $16x^2-8x$ .

What if the numerator and denominator don't have any common factors?

If there are no common factors after complete factorization, then the rational expression is already in simplest form. Not every rational expression can be simplified further!

Can I distribute before factoring instead?

No! For simplifying rational expressions, you want to factor, not distribute. Distributing makes expressions more complex and harder to simplify.

What domain restrictions should I consider?

The denominator cannot equal zero, so $8x ≠ 0$ , which means $x ≠ 0$ . Always check what values make the original denominator zero!

How can I verify my simplified answer is correct?

Pick a test value (like x = 1) and substitute into both the original expression and your simplified answer. They should give the same result: $\frac{2(1)-1}{16(1)^2-8(1)} = \frac{1}{8}$ and $\frac{1}{8(1)} = \frac{1}{8}$ ✓

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