Answered

Select the correct step and statement.



Julian factored the expression $2 x^4+2 x^3-x^2-x$. His work is shown.

At which step did Julian make his first mistake, and which statement describes the mistake?




$2 x^4+2 x^3-x^2-x$

Step 1		$=x\left(2 x^3+2 x^2-x-1\right)$

Step 2		$=x\left[2 x^2(x+1)-1(x-1)\right]$

Step 3		$=x\left(2 x^2-1\right)(x+1)(x-1)$






Statement 1		Julian should have factored $\left(2 x^2-1\right)$ as a difference of squares.

Statement 2		Julian incorrectly applied the distributive property when factoring out -1.

Statement 3		Julian should have factored $2 x$ from all terms instead of $x$.

Statement 4		Julian incorrectly factored $2 x^2$ from the first group of terms.

Verified Answer :

  • Julian's Step 1 is correct: \(2x^4 + 2x^3 - x^2 - x = x(2x^3 + 2x^2 - x - 1)\).
  • In Step 2, Julian incorrectly factors \(-x - 1\) as \(-1(x - 1)\) instead of \(-1(x + 1)\).
  • This is an error in the distributive property when factoring out \(-1\), corresponding to Statement 2.
  • The mistake happens in Step 2, thus the answer is \(\boxed{\text{Step 2, Statement 2}}\).

Explanation

  1. Analyzing Step 1 and Step 2

We are given the expression \(2x^4 + 2x^3 - x^2 - x\) and Julian's attempted factorization steps. We need to identify the first step where he made a mistake and choose the correct statement describing that mistake.

The initial expression is \(2x^4 + 2x^3 - x^2 - x\).
Step 1: \(x(2x^3 + 2x^2 - x - 1)\).
Distributing \(x\) back into the expression: \(x \cdot 2x^3 + x \cdot 2x^2 - x \cdot x - x \cdot 1 = 2x^4 + 2x^3 - x^2 - x\). This step is correct.

Step 2: \(x[2x^2(x + 1) - 1(x - 1)]\).
Let's analyze the polynomial \(2x^3 + 2x^2 - x - 1\) inside the brackets of Step 1.
Using factoring by grouping:
\((2x^3 + 2x^2) + (-x - 1) = 2x^2(x + 1) - 1(x + 1)\).
Comparing this to Julian's result in Step 2, he wrote \(-1(x - 1)\) instead of \(-1(x + 1)\).
When factoring out \(-1\) from \(-x - 1\), the result should be \(-1(x + 1)\), because \(-1 \cdot x = -x\) and \(-1 \cdot 1 = -1\). Julian incorrectly wrote \(-1(x - 1)\), which would expand to \(-x + 1\), not \(-x - 1\). Therefore, Step 2 is the first step where he made a mistake.

  1. Identifying the Mistake and Statement

Based on the analysis, Step 2 is where the error occurred.
The expression inside the brackets in Step 1 is \(2x^3 + 2x^2 - x - 1\).
Factoring the first two terms gives \(2x^2(x+1)\).
Factoring the last two terms gives \(-1(x+1)\).
Combining these, we get \((2x^2 - 1)(x + 1)\).
Julian's Step 2: \(x[2x^2(x + 1) - 1(x - 1)]\).
The error is that Julian wrote \(-1(x-1)\) instead of \(-1(x+1)\).
Looking at the provided statements:
Statement 1: Julian should have factored \((2x^2 - 1)\) as a difference of squares.
Statement 2: Julian incorrectly applied the distributive property when factoring out \(-1\).
Statement 3: Julian should have factored \(2x\) from all terms instead of \(x\).
Statement 4: Julian incorrectly factored \(2x^2\) from the first group of terms.

Statement 2 accurately describes the mistake made: when he factored out \(-1\) from the terms \(-x - 1\), he should have obtained \(-1(x + 1)\), but he incorrectly wrote \(-1(x - 1)\).

  1. Conclusion

Julian's first mistake occurred in Step 2. He incorrectly applied the distributive property when factoring out \(-1\) from the binomial \(-x - 1\), resulting in \(-1(x - 1)\) instead of \(-1(x + 1)\). The correct step should have been \(x[2x^2(x + 1) - 1(x + 1)] = x(2x^2 - 1)(x + 1)\). Therefore, the correct choice is Step 2 and Statement 2. The final answer is \(\boxed{\text{Step 2, Statement 2}}\).

Examples

In algebra, factoring polynomials by grouping requires careful attention to signs, especially when factoring out a negative coefficient. For example, if you are simplifying a expression like \(-x - 5\), factoring out \(-1\) yields \(-1(x + 5)\). If one mistakenly writes \(-1(x - 5)\), the signs will not match the original expression when distributed back. This is a common pitfall when organizing terms and can be checked by expanding the result to ensure it matches the original equation. Master of this skill is essential for solving complex equations and simplifying rational expressions.

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