Simplifying Algebraic Expressions

We are asked to simplify the algebraic expression:

\(\frac{x^2-2 x-63}{x^2+14 x+49}\)

To simplify this fraction, we need to factorize both the numerator and the denominator.

Factorizing the Numerator: \(x^2 - 2x - 63\)

The numerator is a quadratic trinomial of the form \(ax^2 + bx + c\), where \(a=1\), \(b=-2\), and \(c=-63\). We look for two numbers that multiply to \(c\) (\(-63\)) and add up to \(b\) (\(-2\)).

Let the two numbers be \(m\) and \(n\). We need:

• \(m \times n = -63\)
• \(m + n = -2\)

Let's list factors of 63 and see which pair sums to -2:

• 1 and 63 (sum = 64 or -64, product = 63 or -63)
• 3 and 21 (sum = 24 or -24, product = 63 or -63)
• 7 and 9 (sum = 16 or -16, product = 63 or -63)

To get a product of -63, one number must be positive and the other negative. To get a sum of -2, the numbers 7 and 9 are good candidates, with the larger one being negative.

• \(7 \times (-9) = -63\)
• \(7 + (-9) = -2\)

The two numbers are 7 and -9. So, the numerator can be factored as:

\(x^2 - 2x - 63 = (x+7)(x-9)\)

Factorizing the Denominator: \(x^2 + 14x + 49\)

The denominator is also a quadratic trinomial of the form \(ax^2 + bx + c\), where \(a=1\), \(b=14\), and \(c=49\). We look for two numbers that multiply to \(c\) (\(49\)) and add up to \(b\) (\(14\)).

Let the two numbers be \(p\) and \(q\). We need:

• \(p \times q = 49\)
• \(p + q = 14\)

Let's list factors of 49:

• 1 and 49 (sum = 50)
• 7 and 7 (sum = 14)

The two numbers are 7 and 7. So, the denominator can be factored as:

\(x^2 + 14x + 49 = (x+7)(x+7) = (x+7)^2\)

Alternatively, we can recognize the denominator as a perfect square trinomial, which has the form \(a^2 + 2ab + b^2 = (a+b)^2\). Here, \(x^2\) is \(a^2\), so \(a=x\). \(49\) is \(b^2\), so \(b=7\). The middle term is \(2ab = 2(x)(7) = 14x\), which matches the given middle term. Thus, \(x^2 + 14x + 49 = (x+7)^2\).

Simplifying the Expression

Now substitute the factored forms back into the original expression:

\(\frac{x^2-2 x-63}{x^2+14 x+49} = \frac{(x+7)(x-9)}{(x+7)(x+7)}\)

Assuming \(x \neq -7\), we can cancel out the common factor of \((x+7)\) from the numerator and the denominator:

\(\frac{\cancel{(x+7)}(x-9)}{\cancel{(x+7)}(x+7)} = \frac{x-9}{x+7}\)

The simplified expression is \(\frac{x-9}{x+7}\).

Additional Information: Domain of Algebraic Expressions

When simplifying algebraic expressions that involve fractions, it's important to consider the domain of the expression. The domain is the set of all possible values for the variable for which the expression is defined.

An algebraic fraction is undefined when its denominator is equal to zero. In the original expression \(\frac{x^2-2 x-63}{x^2+14 x+49}\), the denominator is \(x^2 + 14x + 49\). Setting the denominator to zero gives:

\(x^2 + 14x + 49 = 0\)

As we factored, this is \((x+7)^2 = 0\). Taking the square root of both sides gives \(x+7 = 0\), which means \(x = -7\).

Therefore, the original expression is undefined when \(x = -7\). While the simplified expression \(\frac{x-9}{x+7}\) also appears to be undefined only at \(x = -7\), it's important to remember that the simplified expression is equal to the original expression only for the values of \(x\) where the original expression is defined. So, the domain of the original expression is all real numbers except \(x = -7\).