The expression \(\frac{{\left( {{x^3} - 1} \right)\left( {{x^2} - 9x + 14} \right)}}{{\left( {{x^2} + x + 1} \right)\left( {{x^2} - 8x + 7} \right)}}\) simplifies to
(x - 2)
Simplifying Algebraic Fractions
The problem asks us to simplify a given rational expression. A rational expression is a fraction where the numerator and the denominator are polynomials. To simplify such expressions, we typically factor both the numerator and the denominator and then cancel out any common factors.
The given expression is:
\[ \frac{{\left( {{x^3} - 1} \right)\left( {{x^2} - 9x + 14} \right)}}{{\left( {{x^2} + x + 1} \right)\left( {{x^2} - 8x + 7} \right)}} \]Step-by-Step Simplification Process
We will factor each polynomial in the numerator and the denominator separately.
Factoring the Numerator:
- The first term in the numerator is \(x^3 - 1\). This is a difference of cubes, which factors as \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\). Here, \(a=x\) and \(b=1\). \[ x^3 - 1 = (x - 1)(x^2 + x \cdot 1 + 1^2) = (x - 1)(x^2 + x + 1) \]
- The second term in the numerator is \(x^2 - 9x + 14\). This is a quadratic trinomial. We look for two numbers that multiply to 14 and add up to -9. These numbers are -2 and -7. \[ x^2 - 9x + 14 = (x - 2)(x - 7) \]
So, the factored numerator is \((x - 1)(x^2 + x + 1)(x - 2)(x - 7)\).
Factoring the Denominator:
- The first term in the denominator is \(x^2 + x + 1\). This quadratic expression does not factor into simpler linear terms with real coefficients.
- The second term in the denominator is \(x^2 - 8x + 7\). This is a quadratic trinomial. We look for two numbers that multiply to 7 and add up to -8. These numbers are -1 and -7. \[ x^2 - 8x + 7 = (x - 1)(x - 7) \]
So, the factored denominator is \((x^2 + x + 1)(x - 1)(x - 7)\).
Putting the Factored Forms Together:
Now we rewrite the original expression using the factored forms:
\[ \frac{(x - 1)(x^2 + x + 1)(x - 2)(x - 7)}{(x^2 + x + 1)(x - 1)(x - 7)} \]Canceling Common Factors:
We can cancel out the factors that appear in both the numerator and the denominator, provided these factors are not equal to zero. The common factors are \((x - 1)\), \((x^2 + x + 1)\), and \((x - 7)\).
Assuming \(x \neq 1\), \(x^2 + x + 1 \neq 0\) (which is true for all real \(x\)), and \(x \neq 7\), we cancel these terms:
\[ \frac{\cancel{(x - 1)}\cancel{(x^2 + x + 1)}(x - 2)\cancel{(x - 7)}}{\cancel{(x^2 + x + 1)}\cancel{(x - 1)}\cancel{(x - 7)}} \]After canceling the common factors, we are left with:
\[ x - 2 \]Thus, the simplified expression is \((x - 2)\).
Revision Table - Algebraic Simplification Concepts
| Concept | Formula/Method | Example |
|---|---|---|
| Difference of Cubes | \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\) | \(x^3 - 1 = (x-1)(x^2+x+1)\) |
| Factoring Quadratic \(ax^2+bx+c\) | Find two numbers \(p, q\) such that \(p \cdot q = ac\) and \(p + q = b\). Rewrite \(bx\) as \(px+qx\) and factor by grouping. For \(x^2+bx+c\), find \(p, q\) such that \(p \cdot q = c\) and \(p + q = b\), then factors are \((x+p)(x+q)\). | \(x^2 - 9x + 14\): Numbers are -2, -7. Factors are \((x-2)(x-7)\). |
| Simplifying Rational Expressions | Factor numerator and denominator, then cancel common factors. | \(\frac{(x+1)(x+2)}{(x+1)(x+3)} = \frac{x+2}{x+3}\) (if \(x \neq -1, x \neq -3\)) |
Additional Information on Rational Expressions
When simplifying rational expressions, it's important to be aware of the values of the variable for which the original expression is undefined. The original expression is undefined when the denominator is zero.
In our case, the denominator is \((x^2 + x + 1)(x^2 - 8x + 7)\). We found that \(x^2 - 8x + 7 = (x - 1)(x - 7)\). The quadratic \(x^2 + x + 1\) is never zero for real values of \(x\) because its discriminant (\(b^2 - 4ac = 1^2 - 4(1)(1) = 1 - 4 = -3\)) is negative. Therefore, the original expression is undefined when \((x - 1)(x - 7) = 0\), which means \(x = 1\) or \(x = 7\).
The simplified expression \((x - 2)\) is defined for all real numbers. However, the simplification step (canceling common factors) is only valid when the canceled factors are non-zero. Thus, the simplified expression is equivalent to the original expression only for values of \(x\) where the original expression is defined (i.e., \(x \neq 1\) and \(x \neq 7\)).
Understanding how to factor different types of polynomials is a fundamental skill for simplifying rational expressions.
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