Question

select the correct answer from each drop-down menu. the quotient of \\(\frac{x^2 + x - 6}{x^2 - 6x + 5} \div \frac{x^2 + 2x - 3}{x^2 - 7x + 10}\\) has \\(\boldsymbol{\square}\\) in the numerator and \\(\boldsymbol{\square}\\) in the denominator.

Explanation:

Step1: Factor each quadratic

Factor \(x^2 + x - 6\): Find two numbers that multiply to \(-6\) and add to \(1\). They are \(3\) and \(-2\), so \(x^2 + x - 6=(x + 3)(x - 2)\).
Factor \(x^2 - 6x + 5\): Find two numbers that multiply to \(5\) and add to \(-6\). They are \(-1\) and \(-5\), so \(x^2 - 6x + 5=(x - 1)(x - 5)\).
Factor \(x^2 + 2x - 3\): Find two numbers that multiply to \(-3\) and add to \(2\). They are \(3\) and \(-1\), so \(x^2 + 2x - 3=(x + 3)(x - 1)\).
Factor \(x^2 - 7x + 10\): Find two numbers that multiply to \(10\) and add to \(-7\). They are \(-2\) and \(-5\), so \(x^2 - 7x + 10=(x - 2)(x - 5)\).

Step2: Rewrite division as multiplication

Dividing by a fraction is multiplying by its reciprocal:
\(\frac{x^2 + x - 6}{x^2 - 6x + 5} \div \frac{x^2 + 2x - 3}{x^2 - 7x + 10}=\frac{x^2 + x - 6}{x^2 - 6x + 5} \times \frac{x^2 - 7x + 10}{x^2 + 2x - 3}\)

Step3: Substitute factored forms

Substitute the factored quadratics:
\(\frac{(x + 3)(x - 2)}{(x - 1)(x - 5)} \times \frac{(x - 2)(x - 5)}{(x + 3)(x - 1)}\)

Step4: Cancel common factors

Cancel \((x + 3)\), \((x - 2)\) (one from numerator and one from denominator? Wait, no: numerator has \((x - 2)\) once in first fraction and once in second, denominator has \((x - 5)\) once in first and once in second, \((x - 1)\) once in first and once in second. Wait, let's list numerators and denominators:
Numerators: \((x + 3)(x - 2) \times (x - 2)(x - 5)=(x + 3)(x - 2)^2(x - 5)\)
Denominators: \((x - 1)(x - 5) \times (x + 3)(x - 1)=(x - 1)^2(x - 5)(x + 3)\)

Cancel \((x + 3)\) (numerator and denominator), \((x - 5)\) (numerator and denominator). Remaining:
Numerator: \((x - 2)^2\)? Wait, no, wait first fraction numerator: \((x + 3)(x - 2)\), second fraction numerator: \((x - 2)(x - 5)\). So total numerator: \((x + 3)(x - 2)(x - 2)(x - 5)\). Denominator: \((x - 1)(x - 5)(x + 3)(x - 1)\). So cancel \((x + 3)\), \((x - 5)\), leaving numerator: \((x - 2)^2\)? Wait, no, \((x - 2)\) is multiplied twice? Wait, no, first fraction numerator: \((x + 3)(x - 2)\), second fraction numerator: \((x - 2)(x - 5)\). So numerator is \((x + 3)(x - 2)(x - 2)(x - 5)\). Denominator is \((x - 1)(x - 5)(x + 3)(x - 1)\). So cancel \((x + 3)\) (both numerator and denominator), \((x - 5)\) (both), then numerator has \((x - 2)^2\)? Wait, no, \((x - 2)\) appears twice in numerator, \((x - 1)\) appears twice in denominator. Wait, no, first fraction denominator: \((x - 1)(x - 5)\), second fraction denominator: \((x + 3)(x - 1)\). So denominator is \((x - 1)(x - 5)(x + 3)(x - 1)=(x - 1)^2(x - 5)(x + 3)\). Numerator is \((x + 3)(x - 2)(x - 2)(x - 5)=(x + 3)(x - 2)^2(x - 5)\). So cancel \((x + 3)\), \((x - 5)\), leaving numerator: \((x - 2)^2\)? Wait, no, that can't be. Wait, maybe I made a mistake in factoring. Wait \(x^2 -7x +10\): factors of 10 that add to -7 are -2 and -5, so \((x - 2)(x - 5)\), correct. \(x^2 +2x -3\): factors of -3 that add to 2 are 3 and -1, so \((x + 3)(x - 1)\), correct. \(x^2 +x -6\): 3 and -2, so \((x + 3)(x - 2)\), correct. \(x^2 -6x +5\): -1 and -5, so \((x - 1)(x - 5)\), correct.

So when we multiply, it's \(\frac{(x + 3)(x - 2)}{(x - 1)(x - 5)} \times \frac{(x - 2)(x - 5)}{(x + 3)(x - 1)}\). Now, multiply numerators: \((x + 3)(x - 2)(x - 2)(x - 5)\). Multiply denominators: \((x - 1)(x - 5)(x + 3)(x - 1)\). Now, cancel common factors:

- \((x + 3)\) is in numerator and denominator: cancel.
- \((x - 5)\) is in numerator and denominator: cancel.
- \((x - 2)\) is in numerator (two times? Wait, first numerator: \((x - 2)\), second numerator: \((x - 2)\): so two \((x - 2)\) in numerator.
- \((x - 1)\) is in denominator (two times: first denominator: \((x - 1)\), second denominator: \((x - 1)\): so two \((x - 1)\) in denominator.

Wait, no, first numerator: \((x + 3)(x - 2)\), second numerator: \((x - 2)(x - 5)\). So numerator has \((x - 2)\) once from first, once from second: total \((x - 2)^2\). Denominator: first denominator \((x - 1)(x - 5)\), second denominator \((x + 3)(x - 1)\): so \((x - 1)^2(x - 5)(x + 3)\). After canceling \((x + 3)\) and \((x - 5)\), numerator is \((x - 2)^2\), denominator is \((x - 1)^2\). Wait, that's different. Wait, maybe I messed up the multiplication. Let's do it again:

First fraction: \(\frac{(x + 3)(x - 2)}{(x - 1)(x - 5)}\)
Second fraction (reciprocal): \(\frac{(x - 2)(x - 5)}{(x + 3)(x - 1)}\)

Multiply numerators: \((x + 3)(x - 2) \times (x - 2)(x - 5) = (x + 3)(x - 2)^2(x - 5)\)
Multiply denominators: \((x - 1)(x - 5) \times (x + 3)(x - 1) = (x - 1)^2(x - 5)(x + 3)\)

Now, cancel \((x + 3)\) (top and bottom), \((x - 5)\) (top and bottom). Left with:

Numerator: \((x - 2)^2\)
Denominator: \((x - 1)^2\)

Wait, but the problem says "the quotient has [blank] in the numerator and [blank] in the denominator". Wait, maybe I made a mistake. Wait, let's check the factoring again. Wait \(x^2 -7x +10\): \((x - 2)(x - 5)\), correct. \(x^2 +2x -3\): \((x + 3)(x - 1)\), correct. \(x^2 +x -6\): \((x + 3)(x - 2)\), correct. \(x^2 -6x +5\): \((x - 1)(x - 5)\), correct.

So after canceling, numerator is \((x - 2)^2\) and denominator is \((x - 1)^2\)? Wait, but maybe the question is expecting factored form before canceling? No, the quotient is after simplifying. Wait, maybe I made a mistake in the number of \((x - 2)\) terms. Wait first fraction numerator: \((x - 2)\), second fraction numerator: \((x - 2)\): so two \((x - 2)\) in numerator. Denominator: two \((x - 1)\) in denominator. So numerator: \((x - 2)^2\), denominator: \((x - 1)^2\). But maybe the question is looking for the simplified numerator and denominator, so numerator is \((x - 2)^2\) or maybe \((x - 2)(x - 2)\) but likely \((x - 2)^2\) and denominator \((x - 1)^2\). Wait, but maybe I messed up the reciprocal. Let's confirm: dividing by \(\frac{A}{B}\) is multiplying by \(\frac{B}{A}\), so yes, that part is correct.

Wait, maybe the original problem has a typo, but assuming the factoring is correct, the numerator after simplifying is \((x - 2)^2\) (or \((x - 2)(x - 2)\)) and denominator is \((x - 1)^2\) (or \((x - 1)(x - 1)\)). But maybe the question expects the numerator to be \((x - 2)(x - 5)\) or something else? No, let's re-express:

Wait, no, wait first fraction numerator: \((x + 3)(x - 2)\), second fraction numerator: \((x - 2)(x - 5)\). So numerator is \((x + 3)(x - 2)(x - 2)(x - 5)\). Denominator is \((x - 1)(x - 5)(x + 3)(x - 1)\). Cancel \((x + 3)\) and \((x - 5)\), so numerator: \((x - 2)^2\), denominator: \((x - 1)^2\). So the numerator is \((x - 2)^2\) (or expanded: \(x^2 -4x +4\)) and denominator is \((x - 1)^2\) (or \(x^2 -2x +1\)). But maybe the question is looking for the factored form before canceling? No, the quotient is the simplified form. Wait, maybe I made a mistake in the number of \((x - 2)\) terms. Wait, first fraction numerator: \((x - 2)\), second fraction numerator: \((x - 2)\): so two \((x - 2)\) in numerator. Denominator: two \((x - 1)\) in denominator. So yes, numerator is \((x - 2)^2\), denominator is \((x - 1)^2\).

But maybe the question is designed to have numerator \((x - 2)(x - 5)\) or something, but no, let's check again. Wait, maybe I flipped the reciprocal. Wait, division of fractions: \(\frac{a}{b} \div \frac{c}{d}=\frac{a}{b} \times \frac{d}{c}\). So here, \(a = x^2 +x -6\), \(b = x^2 -6x +5\), \(c = x^2 +2x -3\), \(d = x^2 -7x +10\). So yes, reciprocal of \(\frac{c}{d}\) is \(\frac{d}{c}\), so multiplication is \(\frac{a}{b} \times \frac{d}{c}\). So that part is correct.

So after factoring and canceling, numerator is \((x - 2)^2\) and denominator is \((x - 1)^2\). So the numerator has \((x - 2)^2\) (or \((x - 2)(x - 2)\)) and denominator has \((x - 1)^2\) (or \((x - 1)(x - 1)\)).

Answer:

Numerator: \((x - 2)^2\) (or \(x^2 - 4x + 4\)), Denominator: \((x - 1)^2\) (or \(x^2 - 2x + 1\))

(Note: If the drop-down options include factored forms, the numerator would be \((x - 2)^2\) and denominator \((x - 1)^2\); if expanded, the numerator is \(x^2 - 4x + 4\) and denominator is \(x^2 - 2x + 1\).)