Solve (x-1)(x+1)(x-2) = -2x²+x³: Comparing Polynomial Forms

Polynomial Factorization with Algebraic Expansion

Solve the following equation:

(x−1)(x+1)(x−2)=−2x2+x3 (x-1)(x+1)(x-2)=-2x^2+x^3

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:04 Let's use the abbreviated multiplication formulas to open the parentheses
00:29 Open parentheses properly, multiply each factor by each factor
00:58 Calculate the multiplications
01:12 Simplify what we can
01:22 Isolate X
01:28 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve the following equation:

(x−1)(x+1)(x−2)=−2x2+x3 (x-1)(x+1)(x-2)=-2x^2+x^3

2

Step-by-step solution

Let's examine the given equation:

(x−1)(x+1)(x−2)=−2x2+x3 (x-1)(x+1)(x-2)=-2x^2+x^3 First, let's simplify the equation, using the perfect square difference formula:

(a+b)(a−b)=a2−b2 (a+b)(a-b)=a^2-b^2 and the expanded distribution law,

We'll start by opening the first pair of parentheses from the left which is in the left side using the perfect square difference formula mentioned, we'll put the result in new parentheses (since the entire expression is multiplied by the expression in the unopened parentheses) then we'll simplify the expression in the parentheses:

(x−1)(x+1)(x−2)=−2x2+x3↓(x2−12)(x−2)=−2x2+x3(x2−1)(x−2)=−2x2+x3 (x-1)(x+1)(x-2)=-2x^2+x^3 \\ \downarrow\\ \textcolor{blue}{(}x^2-1^2\textcolor{blue}{)}(x-2)=-2x^2+x^3\\ \textcolor{blue}{(}x^2-1\textcolor{blue}{)}(x-2)=-2x^2+x^3\\ We'll continue using the expanded distribution law and open the parentheses on the left side, then we'll move terms and combine like terms:

(x2−1)(x−2)=−2x2+x3↓x3−2x2−x+2=−2x2+x3x3−2x2−x+2+2x2−x3=0−x+2=0−x=−2/⋅(−1)x=2 (x^2-1)(x-2)=-2x^2+x^3\\ \downarrow\\ x^3-2x^2-x+2=-2x^2+x^3\\ x^3-2x^2-x+2+2x^2-x^3=0\\ -x+2=0\\ -x=-2\hspace{6pt}\text{/}\cdot(-1)\\ \boxed{x=2}

Let's summarize the equation solving steps:

(x−1)(x+1)(x−2)=−2x2+x3↓(x2−1)(x−2)=−2x2+x3↓x3−2x2−x+2=−2x2+x3x=2 (x-1)(x+1)(x-2)=-2x^2+x^3 \\ \downarrow\\ \textcolor{blue}{(}x^2-1\textcolor{blue}{)}(x-2)=-2x^2+x^3\\ \downarrow\\ x^3-2x^2-x+2=-2x^2+x^3\\ \boxed{x=2} Therefore, the correct answer is answer C.

3

Final Answer

x=2 x=2

Key Points to Remember

Essential concepts to master this topic
  • Rule: Use difference of squares formula (a+b)(a−b)=a2−b2 (a+b)(a-b) = a^2-b^2
  • Technique: Expand (x−1)(x+1)=x2−1 (x-1)(x+1) = x^2-1 before distributing
  • Check: Substitute x = 2: (1)(3)(0)=−8+8=0 (1)(3)(0) = -8+8 = 0 ✓

Common Mistakes

Avoid these frequent errors
  • Expanding all three factors at once
    Don't try to multiply (x−1)(x+1)(x−2) (x-1)(x+1)(x-2) all together immediately = messy algebra! This creates unnecessary complexity with six terms to track. Always group factors strategically using special formulas like difference of squares first.

Practice Quiz

Test your knowledge with interactive questions

\( x^2+6x+9=0 \)

What is the value of X?

FAQ

Everything you need to know about this question

Why can't I just expand all three factors directly?

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You could, but it's much more work! Using the difference of squares formula (x−1)(x+1)=x2−1 (x-1)(x+1) = x^2-1 saves time and reduces errors.

How do I know which factors to group first?

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Look for special patterns! Here, (x−1)(x+1) (x-1)(x+1) fits the difference of squares formula perfectly, so group these first.

What if I get confused with all the algebra?

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Take it step by step! First use the special formula, then distribute, then move all terms to one side. Don't rush - each step should be clear.

How do I check if x = 2 is really the answer?

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Substitute x = 2 into the original equation: Left side = (1)(3)(0)=0 (1)(3)(0) = 0 , Right side = −8+8=0 -8+8 = 0 . Both equal 0, so x = 2 is correct!

Why does the equation simplify to just -x + 2 = 0?

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After expanding and moving terms, most cancel out! The x3 x^3 and −2x2 -2x^2 terms appear on both sides, so they subtract to zero, leaving only the linear term.

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