Question

1. simplify each of the following rational expressions, and state the non - permissible value(s). (12 m

a. \\(\frac{x^2 - 4}{x^2 - 5x + 6}\\)
\\(x^2 - 5x + 6\
eq0\\)
\\((x - 6)(x - 1)\
eq0\\)
\\(x - 6\
eq0\\) \\(x - 1\
eq0\\)
\\(\boxed{x\
eq6}\\) \\(\boxed{x\
eq1}\\)
\\(=\frac{(x - 2)(x + 2)}{(x - 6)(x - 1)}
=\\)
b. \\(\frac{3x - 9}{x^2 + 2x - 15}\\)
\\(x^2 + 2x - 15\
eq0\\)
\\((x - 3)(x + 5)\
eq0\\)
\\(x - 3\
eq0\\) \\(x + 5\
eq0\\)
\\(\boxed{x\
eq3}\\) \\(\boxed{x\
eq - 5}\\)
\\(=\frac{3(x - 3)}{(x - 3)(x + 5)}
=\boxed{\frac{3}{x + 5}}\\)
c. \\(\frac{x^2 - 4x - 21}{x^2 - 9}\\)
d. \\(\frac{6x - 3}{1 - 4x^2}\\)

Explanation:

Step1: Factor Numerator and Denominator

For part a:
Numerator: \(x^2 - 4=(x - 2)(x + 2)\) (difference of squares).
Denominator: \(x^2 - 5x + 6=(x - 2)(x - 3)\) (factoring quadratic). Wait, the original work had a mistake here! The correct factorization of \(x^2 - 5x + 6\) is \((x - 2)(x - 3)\), not \((x - 6)(x - 1)\). Let's correct that.

Non - permissible values: Denominator \(
eq0\), so \((x - 2)(x - 3)
eq0\Rightarrow x
eq2,x
eq3\).

Step2: Simplify the Rational Expression

Cancel common factors: \(\frac{(x - 2)(x + 2)}{(x - 2)(x - 3)}=\frac{x + 2}{x - 3}\) (for \(x
eq2,3\)).

Step1 (Part b - Correcting Original Work):

Original denominator factoring: \(x^2+2x - 15=(x + 5)(x - 3)\) (correct). Numerator: \(3x - 9 = 3(x - 3)\).

Non - permissible values: \((x + 5)(x - 3)
eq0\Rightarrow x
eq - 5,x
eq3\).

Step2 (Part b - Simplify):

Cancel \((x - 3)\) (since \(x
eq3\), it's a valid cancellation): \(\frac{3(x - 3)}{(x + 5)(x - 3)}=\frac{3}{x + 5}\) (for \(x
eq - 5,3\)).

Step1 (Part c):

Numerator: \(x^2-4x - 21=(x - 7)(x + 3)\) (factoring quadratic: find two numbers with product \(-21\) and sum \(-4\), which are \(-7\) and \(3\)).
Denominator: \(x^2 - 9=(x - 3)(x + 3)\) (difference of squares).

Non - permissible values: \((x - 3)(x + 3)
eq0\Rightarrow x
eq3,x
eq - 3\).

Step2 (Part c - Simplify):

Cancel \((x + 3)\) (since \(x
eq - 3\)): \(\frac{(x - 7)(x + 3)}{(x - 3)(x + 3)}=\frac{x - 7}{x - 3}\) (for \(x
eq3,-3\)).

Step1 (Part d):

Numerator: \(6x - 3 = 3(2x - 1)\).
Denominator: \(1-4x^2=(1 - 2x)(1 + 2x)=-(2x - 1)(1 + 2x)\) (difference of squares: \(a^2 - b^2=(a - b)(a + b)\), here \(a = 1,b = 2x\)).

Non - permissible values: \(1-4x^2
eq0\Rightarrow(1 - 2x)(1 + 2x)
eq0\Rightarrow x
eq\frac{1}{2},x
eq-\frac{1}{2}\).

Step2 (Part d - Simplify):

Rewrite numerator and denominator: \(\frac{3(2x - 1)}{-(2x - 1)(1 + 2x)}\). Cancel \((2x - 1)\) (since \(x
eq\frac{1}{2}\)): \(\frac{3(2x - 1)}{-(2x - 1)(1 + 2x)}=-\frac{3}{1 + 2x}\) (for \(x
eq\frac{1}{2},-\frac{1}{2}\)).

Answer:

s (with Corrections for Part a and Completing Others):
a. Simplified: \(\boldsymbol{\frac{x + 2}{x - 3}}\), Non - permissible: \(x
eq2,3\)
b. Simplified: \(\boldsymbol{\frac{3}{x + 5}}\), Non - permissible: \(x
eq - 5,3\)
c. Simplified: \(\boldsymbol{\frac{x - 7}{x - 3}}\), Non - permissible: \(x
eq3,-3\)
d. Simplified: \(\boldsymbol{-\frac{3}{2x + 1}}\) (or \(-\frac{3}{1 + 2x}\)), Non - permissible: \(x
eq\frac{1}{2},-\frac{1}{2}\)