Solve (x+1)(x-1)(x+1) = x^3 + x^2: Cubic Equation with Repeated Factor

Cubic Equations with Repeated Factors

Solve the following problem:

(x+1)(x1)(x+1)=x2+x3 (x+1)(x-1)(x+1)=x^2+x^3

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:09 Extract a common factor from the parentheses
00:16 Reduce what's possible
00:32 Use the shortened multiplication formulas to open the parentheses
00:40 Reduce what's possible
00:44 We got an illogical expression
00:49 Let's check if we divided by 0, which is why we got an illogical expression
00:52 Let's check if X equals minus 1, meaning we divided by 0
00:57 Let's substitute -1 in the exercise and solve
01:14 Now we got a logical expression, so this is the solution for X
01:18 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 problem:

(x+1)(x1)(x+1)=x2+x3 (x+1)(x-1)(x+1)=x^2+x^3

2

Step-by-step solution

Let's examine the given equation:

(x+1)(x1)(x+1)=x2+x3 (x+1)(x-1)(x+1)=x^2+x^3

Begin by opening the second and third pairs of parentheses from the left (marked with an underline below) On the left side we can apply the difference of squares formula:

(a+b)(ab)=a2b2 (a+b)(a-b)=a^2-b^2 ,

Place the result inside of new parentheses (since the resulting expression in its entirety is multiplied by an expression that is enclosed by these parentheses) then we'll proceed to simplify the expression in the resulting parentheses:

(x+1)(x1)(x+1)=x2+x3(x+1)(x212)=x2+x3(x+1)(x21)=x2+x3 (x+1)\underline{(x-1)(x+1)}=x^2+x^3 \\ \downarrow\\ (x+1)\underline{\textcolor{blue}{(}x^2-1^2\textcolor{blue}{)}}=x^2+x^3\\ (x+1)\underline{\textcolor{blue}{(}x^2-1\textcolor{blue}{)}}=x^2+x^3\\ Continue to use the expanded distributive law again and open the parentheses on the left side. Proceed to move and combine like terms:

(x+1)(x21)=x2+x3x3x+x21=x2+x3x=1/(1)x=1 (x+1)(x^2-1)=x^2+x^3\\ \downarrow\\ x^3-x+x^2-1=x^2+x^3\\ -x =1\hspace{6pt}\text{/}\cdot(-1)\\ \downarrow\\ \boxed{x=-1}

In the final step solve the first-degree equation that we obtained,

Therefore the correct answer is answer A.

3

Final Answer

x=1 x=-1

Key Points to Remember

Essential concepts to master this topic
  • Factoring: Use difference of squares formula (a+b)(ab)=a2b2 (a+b)(a-b) = a^2-b^2
  • Technique: (x+1)(x1)(x+1)=(x+1)(x21) (x+1)(x-1)(x+1) = (x+1)(x^2-1) then distribute
  • Check: Substitute x = -1: both sides equal 0 ✓

Common Mistakes

Avoid these frequent errors
  • Expanding all three factors individually
    Don't expand (x+1)(x-1)(x+1) by multiplying each factor separately = messy algebra with many terms! This creates unnecessary complexity and increases error chances. Always look for patterns like difference of squares first to simplify before expanding.

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 do I use the difference of squares formula here?

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The pattern (x+1)(x1) (x+1)(x-1) is a perfect match for difference of squares! This gives us x21 x^2 - 1 instantly, which is much simpler than expanding manually.

Can I solve this by expanding everything first?

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Yes, but it's much harder! You'd get x3+x2x1=x3+x2 x^3 + x^2 - x - 1 = x^3 + x^2 , which still simplifies to the same result. The pattern recognition saves time and reduces errors.

What if I missed the repeated (x+1) factor?

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The equation has two (x+1) factors, so it's really (x+1)2(x1) (x+1)^2(x-1) . Either way works - use difference of squares on any two factors that fit the pattern!

How do I know x = -1 is the only solution?

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After simplifying, we get x=1 -x = 1 , which is a linear equation with exactly one solution. Unlike quadratic equations, linear equations always have exactly one answer.

Why don't the other answer choices work?

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Test them! For x = 0: left side = (-1)(1)(-1) = 1, right side = 0. For x = 1: left side = (2)(0)(2) = 0, right side = 2. Only x = -1 makes both sides equal.

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