Question

1. the simplified form of the rational expression \\(\frac{4x}{3x^2 - 7x + 4} + \frac{3x + 2}{x^2 - 4x + 3}\\) is

a. \\(\frac{7x + 2}{(4x - 7)(x - 1)}, x \
eq \frac{4}{3}, 1, 3\\)
b. \\(\frac{x - 2}{(2x - 1)(x - 1)}, x \
eq \frac{4}{3}, 1, 3\\)
c. \\(\frac{13x^2 - 18x - 8}{(3x - 4)(x - 1)(x - 3)}, x \
eq \frac{4}{3}, 1, 3\\)
d. \\(\frac{-5x^2 - 6x + 8}{(3x - 4)(x - 1)(x - 3)}, x \
eq \frac{4}{3}, 1, 3\\)

1. the simplified form of the rational expression \\(\frac{3x^2 + x - 2}{2x^2 - 9x - 5} \cdot \frac{3x^2 - 14x - 5}{x^2 - 1} \div \frac{9x^2 - 3x - 2}{2x^2 - 7x - 4}\\) is

a. \\(\frac{(x - 4)(3x - 1)}{(x - 1)(2x + 1)}, x \
eq -\frac{1}{2}, -\frac{1}{3}, \pm 1, \frac{2}{3}, 4, 5\\)
b. \\(\frac{x - 4}{x - 1}, x \
eq -\frac{1}{2}, -\frac{1}{3}, \pm 1, \frac{2}{3}, 4, 5\\)
c. \\(\frac{(3x - 2)(3x - 1)}{(2x + 1)(x - 1)(x - 4)}, x \
eq -\frac{1}{2}, -\frac{1}{3}, \pm 1, \frac{2}{3}, 4, 5\\)
d. \\(\frac{(3x - 2)^2(3x - 1)^2}{(2x + 1)^2(x - 1)(x - 4)}, x \
eq -\frac{1}{2}, -\frac{1}{3}, \pm 1, \frac{2}{3}, 4, 5\\)

Explanation:

Question 3

Step 1: Factor the denominators

First, factor \(3x^2 - 7x + 4\). We need two numbers that multiply to \(3\times4 = 12\) and add to \(-7\). The numbers are \(-3\) and \(-4\). So,

$$
LATEXBLOCK0
$$

Next, factor \(x^2 - 4x + 3\). We need two numbers that multiply to \(3\) and add to \(-4\). The numbers are \(-1\) and \(-3\). So,

$$ x^2 - 4x + 3=(x - 1)(x - 3) $$

Step 2: Find the least common denominator (LCD)

The denominators are \((3x - 4)(x - 1)\) and \((x - 1)(x - 3)\). The LCD is \((3x - 4)(x - 1)(x - 3)\).

Step 3: Rewrite the fractions with the LCD

Rewrite \(\frac{4x}{(3x - 4)(x - 1)}\) with denominator \((3x - 4)(x - 1)(x - 3)\):

$$ \frac{4x}{(3x - 4)(x - 1)}=\frac{4x(x - 3)}{(3x - 4)(x - 1)(x - 3)} $$

Rewrite \(\frac{3x + 2}{(x - 1)(x - 3)}\) with denominator \((3x - 4)(x - 1)(x - 3)\):

$$ \frac{3x + 2}{(x - 1)(x - 3)}=\frac{(3x + 2)(3x - 4)}{(3x - 4)(x - 1)(x - 3)} $$

Step 4: Add the fractions

$$
LATEXBLOCK1
$$

Expand the numerators:

$$
LATEXBLOCK2
$$

Add the expanded numerators:

$$
LATEXBLOCK3
$$

So the simplified form is \(\frac{13x^2 - 18x - 8}{(3x - 4)(x - 1)(x - 3)}, x
eq\frac{4}{3},1,3\)

Step 1: Factor all the quadratic expressions

• Factor \(3x^2 + x - 2\): We need two numbers that multiply to \(3\times(-2)= -6\) and add to \(1\). The numbers are \(3\) and \(-2\). So,

$$
LATEXBLOCK0
$$

• Factor \(2x^2 - 9x - 5\): We need two numbers that multiply to \(2\times(-5)= -10\) and add to \(-9\). The numbers are \(-10\) and \(1\). So,

$$
LATEXBLOCK1
$$

• Factor \(3x^2 - 14x - 5\): We need two numbers that multiply to \(3\times(-5)= -15\) and add to \(-14\). The numbers are \(-15\) and \(1\). So,

$$
LATEXBLOCK2
$$

• Factor \(x^2 - 1=(x - 1)(x + 1)\) (difference of squares)
• Factor \(9x^2 - 3x - 2\): We need two numbers that multiply to \(9\times(-2)= -18\) and add to \(-3\). The numbers are \(-6\) and \(3\). So,

$$
LATEXBLOCK3
$$

• Factor \(2x^2 - 7x - 4\): We need two numbers that multiply to \(2\times(-4)= -8\) and add to \(-7\). The numbers are \(-8\) and \(1\). So,

$$
LATEXBLOCK4
$$

Step 2: Rewrite the division as multiplication by reciprocal

The expression \(\frac{3x^2 + x - 2}{2x^2 - 9x - 5}\cdot\frac{3x^2 - 14x - 5}{x^2 - 1}\div\frac{9x^2 - 3x - 2}{2x^2 - 7x - 4}\) becomes:

$$ \frac{3x^2 + x - 2}{2x^2 - 9x - 5}\cdot\frac{3x^2 - 14x - 5}{x^2 - 1}\cdot\frac{2x^2 - 7x - 4}{9x^2 - 3x - 2} $$

Step 3: Substitute the factored forms

$$ \frac{(3x - 2)(x + 1)}{(x - 5)(2x + 1)}\cdot\frac{(x - 5)(3x + 1)}{(x - 1)(x + 1)}\cdot\frac{(x - 4)(2x + 1)}{(3x - 2)(3x + 1)} $$

Step 4: Cancel out common factors

• Cancel \((3x - 2)\) from numerator and denominator.
• Cancel \((x + 1)\) from numerator and denominator.
• Cancel \((x - 5)\) from numerator and denominator.
• Cancel \((3x + 1)\) from numerator and denominator.
• Cancel \((2x + 1)\) from numerator and denominator.

After canceling, we are left with \(\frac{x - 4}{x - 1}\)

Now, find the values of \(x\) for which the original expression is undefined. These are the values that make any denominator zero:

• For \(2x^2 - 9x - 5=(x - 5)(2x + 1)\), \(x = 5\) or \(x=-\frac{1}{2}\)
• For \(x^2 - 1=(x - 1)(x + 1)\), \(x = 1\) or \(x=-1\)
• For \(9x^2 - 3x - 2=(3x - 2)(3x + 1)\), \(x=\frac{2}{3}\) or \(x =-\frac{1}{3}\)
• For \(2x^2 - 7x - 4=(x - 4)(2x + 1)\), \(x = 4\) or \(x=-\frac{1}{2}\) (we already have \(x =-\frac{1}{2}\))

So the simplified form is \(\frac{x - 4}{x - 1}, x
eq-\frac{1}{2},-\frac{1}{3},\pm1,\frac{2}{3},4,5\)

Answer:

C. \(\frac{13x^2 - 18x - 8}{(3x - 4)(x - 1)(x - 3)}, x
eq\frac{4}{3},1,3\)

Question 4