Question

Find the real solutions of each equation by factoring. $$ x^{3}+x^{2}-x-1=0 $$

Short answer

The real solutions are x = -1 and x = 1.

Step-by-step solution

- Group the terms

Group the terms in pairs to make the equation easier to factor:o x^3 + x^2 - x - 1 = 0o (x^3 + x^2) + (-x - 1) = 0.

- Factor out the common factors

Factor out the common factor from each group:o x^2(x + 1) - 1(x + 1) = 0.

- Factor by grouping

Observe that (x + 1) is a common factor in each group. Factor out (x + 1):o (x + 1)(x^2 - 1) = 0.

- Factor the quadratic expression

Notice that x^2 - 1 is a difference of squares and can be factored further:o (x + 1)(x + 1)(x - 1) = 0.oro (x + 1)^2 (x - 1) = 0.

- Set each factor to zero and solve for x

Set each factor equal to zero and solve for x:o (x + 1) = 0 or (x - 1) = 0.Therefore, x = -1 or x = 1.

Key concepts

Factoring

Factoring is a method used to break down polynomial equations into simpler components called factors. This technique is essential for solving polynomial equations because it simplifies the problem and allows us to find the roots (solutions) more easily.
In the given exercise, we start with the equation ewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewline\((x^3 + x^2 - x - 1 = 0)\).
By grouping the terms in pairs and factoring out the common factors, we can then solve the equation step by step. Factoring often involves recognizing patterns and using common techniques, like factoring out the greatest common divisor or factoring by grouping.

Grouping

Grouping is a powerful method to simplify polynomial equations by arranging terms into pairs or groups that can be factored more easily. In our exercise, we initially group the terms:
ewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewline\((x^3 + x^2) + (-x - 1) = 0\).
Once grouped, each pair can be factored individually:
- From \((x^3 + x^2)\), we factor out \(x^2\), giving us \(x^2(x + 1)\).
- From \((-x - 1)\), we factor out \(-1\), giving us \(-1(x + 1)\).
By grouping, we identified a common factor (x + 1), which then simplifies the polynomial.

Difference of Squares

The difference of squares is a special factoring technique used when a polynomial can be written in the form \(a^2 - b^2\). This identity states that \(a^2 - b^2 = (a + b)(a - b)\). In our solution, we encounter a quadratic expression \(x^2 - 1\) during step 4, which is a difference of squares:
\(x^2 - 1 = (x)^2 - (1)^2\).
By applying the difference of squares identity, we factor this expression as: \((x + 1)(x - 1)\).
This step further simplifies our polynomial equation.

Quadratic Expression

A quadratic expression is a polynomial of degree two, usually in the form \(ax^2 + bx + c\). In the given exercise, we worked with a quadratic expression in step 4 - specifically \(x^2 - 1\). Though simple, quadratic expressions can be factored, solved using the quadratic formula, or recognized as a part of a special pattern like the difference of squares.
When we expressed our solution in the form:\((x + 1)(x + 1)(x - 1) = 0\), we were able to solve for \(x\) by setting each factor to zero.
Ultimately, solving quadratic expressions efficiently aids in the overall process of breaking down and solving polynomial equations.