Question

10 what is the quotient? state any restrictions. \\(\frac{x^2 - 16}{x^2 + 5x + 6} \div \frac{x^2 + 5x + 4}{x^2 - 2x - 8}\\) \\(\boldsymbol{a}\\) \\(\frac{(x + 4)^2(x + 1)}{(x + 2)^2(x + 3)}; x \
eq -3, -2, 4\\) \\(\boldsymbol{b}\\) \\(\frac{(x - 4)^2}{(x + 3)(x + 1)}; x \
eq -3, -1\\) \\(\boldsymbol{c}\\) \\(\frac{(x - 4)^2}{(x + 3)(x + 1)}; x \
eq -4, -3, -2, -1, 4\\) \\(\boldsymbol{d}\\) \\(\frac{1}{(x + 3)(x + 1)}; x \
eq -4, -3, -2, -1, 4\\)

Explanation:

Step1: Factor all polynomials

Factor \(x^2 - 16\) as \((x - 4)(x + 4)\) (difference of squares: \(a^2 - b^2=(a - b)(a + b)\)).
Factor \(x^2 + 5x + 6\) as \((x + 2)(x + 3)\) (find two numbers that multiply to 6 and add to 5: 2 and 3).
Factor \(x^2 + 5x + 4\) as \((x + 1)(x + 4)\) (numbers: 1 and 4).
Factor \(x^2 - 2x - 8\) as \((x - 4)(x + 2)\) (numbers: -4 and 2, since \(-4\times2=-8\) and \(-4 + 2=-2\)).

Step2: Rewrite division as multiplication

Dividing by a fraction is multiplying by its reciprocal:
\(\frac{x^2 - 16}{x^2 + 5x + 6}\div\frac{x^2 + 5x + 4}{x^2 - 2x - 8}=\frac{(x - 4)(x + 4)}{(x + 2)(x + 3)}\times\frac{(x - 4)(x + 2)}{(x + 1)(x + 4)}\)

Step3: Cancel common factors

Cancel \((x + 4)\), \((x + 2)\) from numerator and denominator:

• Cancel \((x + 4)\) (one in numerator, one in denominator).
• Cancel \((x + 2)\) (one in numerator, one in denominator).

Left with \(\frac{(x - 4)(x - 4)}{(x + 3)(x + 1)}=\frac{(x - 4)^2}{(x + 3)(x + 1)}\)

Step4: Find restrictions

Denominators cannot be zero:

• From \(x^2 + 5x + 6=(x + 2)(x + 3)\): \(x

eq - 2,-3\).

• From \(x^2 + 5x + 4=(x + 1)(x + 4)\): \(x

eq - 1,-4\).

• From \(x^2 - 2x - 8=(x - 4)(x + 2)\): \(x

eq 4,-2\) (already considered \(-2\) from before).

• Also, check original numerators/denominators for additional restrictions (but we already covered all factors). Wait, wait—wait, when we canceled, we must ensure all original denominators (and the reciprocal’s denominator) are non - zero. So:
• \(x^2 + 5x + 6

eq0\Rightarrow x
eq - 3,-2\).

• \(x^2 + 5x + 4

eq0\Rightarrow x
eq - 1,-4\).

• \(x^2 - 2x - 8

eq0\Rightarrow x
eq 4,-2\) (but \(-2\) is already in the first set).

• Also, check if any canceled factors make numerator zero? No, restrictions are from denominators. Wait, but when we canceled \((x - 4)\), but \((x - 4)\) is in the numerator—wait, no, restrictions are from where the denominator was zero before canceling. Wait, let's re - evaluate:

Original denominators:

1. \(x^2 + 5x + 6=(x + 2)(x + 3)\): \(x

eq - 3,-2\).

1. \(x^2 + 5x + 4=(x + 1)(x + 4)\): \(x

eq - 1,-4\).

1. \(x^2 - 2x - 8=(x - 4)(x + 2)\): \(x

eq 4,-2\) (so \(x
eq 4\) is new, \(-2\) already listed).

So combined restrictions: \(x
eq - 4,-3,-2,-1,4\) (since \(x = - 4\) makes \(x^2 + 5x + 4 = 0\), \(x = 4\) makes \(x^2 - 2x - 8 = 0\), \(x=-3\) makes \(x^2 + 5x + 6 = 0\), \(x=-2\) makes \(x^2 + 5x + 6\) and \(x^2 - 2x - 8 = 0\), \(x=-1\) makes \(x^2 + 5x + 4 = 0\)).

Answer:

C. \(\frac{(x - 4)^2}{(x + 3)(x + 1)}; x
eq - 4, - 3, - 2, - 1, 4\)