Question

perform the indicated operation and simplify the result. leave your answer in factored form.
\\(\frac{x^2 - 3x - 28}{x^2 + 2x - 63} cdot \frac{x^2 + 7x - 18}{x^2 + 13x + 36}\\)
\\(\frac{x^2 - 3x - 28}{x^2 + 2x - 63} cdot \frac{x^2 + 7x - 18}{x^2 + 13x + 36} = square\\)
(simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Factor each quadratic

Factor \(x^2 - 3x - 28\): Find two numbers that multiply to \(-28\) and add to \(-3\). They are \(-7\) and \(4\). So, \(x^2 - 3x - 28=(x - 7)(x + 4)\).
Factor \(x^2 + 2x - 63\): Find two numbers that multiply to \(-63\) and add to \(2\). They are \(9\) and \(-7\). So, \(x^2 + 2x - 63=(x + 9)(x - 7)\).
Factor \(x^2 + 7x - 18\): Find two numbers that multiply to \(-18\) and add to \(7\). They are \(9\) and \(-2\). So, \(x^2 + 7x - 18=(x + 9)(x - 2)\).
Factor \(x^2 + 13x + 36\): Find two numbers that multiply to \(36\) and add to \(13\). They are \(9\) and \(4\). So, \(x^2 + 13x + 36=(x + 9)(x + 4)\).

Step2: Substitute factored forms

Substitute into the original expression:
\[ \frac{(x - 7)(x + 4)}{(x + 9)(x - 7)} \cdot \frac{(x + 9)(x - 2)}{(x + 9)(x + 4)} \]

Step3: Cancel common factors

Cancel \((x - 7)\), \((x + 4)\), and one \((x + 9)\) from numerator and denominator:
After canceling, we have \(\frac{x - 2}{x + 9}\). Wait, no—wait, let's check again. Wait, original factoring:
Wait, \(x^2 + 13x + 36=(x + 9)(x + 4)\) (correct, \(9+4=13\), \(9\times4=36\)). \(x^2 + 7x - 18=(x + 9)(x - 2)\) (correct, \(9-2=7\), \(9\times(-2)=-18\)). \(x^2 - 3x -28=(x -7)(x +4)\) (correct, \(-7+4=-3\), \(-7\times4=-28\)). \(x^2 +2x -63=(x +9)(x -7)\) (correct, \(9-7=2\), \(9\times(-7)=-63\)).

So substituting:
\[ \frac{(x - 7)(x + 4)}{(x + 9)(x - 7)} \cdot \frac{(x + 9)(x - 2)}{(x + 9)(x + 4)} \]

Cancel \((x - 7)\) (numerator of first, denominator of first), \((x + 4)\) (numerator of first, denominator of second), and one \((x + 9)\) (denominator of first, numerator of second). Wait, first fraction's denominator: \((x + 9)(x - 7)\), second fraction's numerator: \((x + 9)(x - 2)\), second fraction's denominator: \((x + 9)(x + 4)\).

So cancel \((x - 7)\) (first numerator and first denominator), \((x + 4)\) (first numerator and second denominator), and one \((x + 9)\) (first denominator and second numerator, and second denominator has \((x + 9)\)). Wait, let's write numerators: \((x -7)(x +4) \times (x +9)(x -2)\). Denominators: \((x +9)(x -7) \times (x +9)(x +4)\).

So numerator factors: \((x -7)\), \((x +4)\), \((x +9)\), \((x -2)\).
Denominator factors: \((x +9)\), \((x -7)\), \((x +9)\), \((x +4)\).

Cancel \((x -7)\), \((x +4)\), one \((x +9)\). Remaining numerator: \((x -2)\). Remaining denominator: \((x +9)\). Wait, no—wait, denominator has two \((x +9)\) (from first denominator: \((x +9)\) and second denominator: \((x +9)\)), numerator has one \((x +9)\). So cancel one \((x +9)\) from numerator and one from denominator. Then:

Numerator: \((x -7)\cancel{(x +4)} \times \cancel{(x +9)}(x -2)\)
Denominator: \(\cancel{(x +9)}(x -7) \times (x +9)\cancel{(x +4)}\)

So after canceling, numerator: \((x -2)\), denominator: \((x +9)\). Wait, but wait, is that right? Wait, let's re-express:

Original expression:
\[ \frac{(x - 7)(x + 4)}{(x + 9)(x - 7)} \times \frac{(x + 9)(x - 2)}{(x + 9)(x + 4)} = \frac{(x - 7)(x + 4)(x + 9)(x - 2)}{(x + 9)(x - 7)(x + 9)(x + 4)} \]

Now, cancel \((x -7)\) (top and bottom), \((x +4)\) (top and bottom), one \((x +9)\) (top and bottom). So remaining: \(\frac{x - 2}{x + 9}\). Wait, but wait, let's check with x=0: original expression:
First fraction: \(\frac{0 -0 -28}{0 +0 -63}=\frac{-28}{-63}=\frac{4}{9}\). Second fraction: \(\frac{0 +0 -18}{0 +0 +36}=\frac{-18}{36}=-\frac{1}{2}\). Multiply: \(\frac{4}{9} \times (-\frac{1}{2})=-\frac{2}{9}\). Now, \(\frac{x -2}{x +9}\) at x=0: \(\frac{-2}{9}\), which matches. So that's correct. Wait, but wait, did I make a mistake in factoring? Let's recheck each quadratic:

- \(x^2 -3x -28\): discriminant \(9 + 112=121\), square root 11. So roots \(\frac{3\pm11}{2}\), so \(x=7\) or \(x=-4\). So factors \((x -7)(x +4)\) (correct).
- \(x^2 +2x -63\): discriminant \(4 + 252=256\), square root 16. Roots \(\frac{-2\pm16}{2}\), so \(x=7\) or \(x=-9\). Wait, wait! Oh no! I made a mistake here. \(x^2 +2x -63\): let's solve \(x^2 +2x -63=0\). \(x=\frac{-2\pm\sqrt{4 + 252}}{2}=\frac{-2\pm16}{2}\). So \(x=\frac{14}{2}=7\) or \(x=\frac{-18}{2}=-9\). So factors are \((x -7)(x +9)\)? Wait, no: if root is 7, then (x -7); root is -9, then (x +9). So \(x^2 +2x -63=(x -7)(x +9)\)? Wait, no: \((x -7)(x +9)=x^2 +9x -7x -63=x^2 +2x -63\). Yes! I had a mistake earlier. I factored it as \((x +9)(x -7)\), which is correct, but when I thought about the roots, I confused the sign. So \((x +9)(x -7)\) is correct (since \(-7 +9=2\), \(-7\times9=-63\)). So that part was correct.

- \(x^2 +7x -18\): discriminant \(49 + 72=121\), square root 11. Roots \(\frac{-7\pm11}{2}\), so \(x=2\) or \(x=-9\). So factors \((x -2)(x +9)\) (correct, since \(-2 +9=7\), \(-2\times9=-18\)). Wait, earlier I had \((x +9)(x -2)\), which is the same as \((x -2)(x +9)\), so that's correct.

- \(x^2 +13x +36\): discriminant \(169 - 144=25\), square root 5. Roots \(\frac{-13\pm5}{2}\), so \(x=-4\) or \(x=-9\). So factors \((x +4)(x +9)\) (correct, \(4 +9=13\), \(4\times9=36\)).

So going back, the factoring was correct. So the original expression:

\[ \frac{(x -7)(x +4)}{(x +9)(x -7)} \times \frac{(x +9)(x -2)}{(x +9)(x +4)} \]

Now, cancel \((x -7)\) (numerator of first, denominator of first), \((x +4)\) (numerator of first, denominator of second), and one \((x +9)\) (denominator of first, numerator of second). Wait, denominator of first is \((x +9)(x -7)\), numerator of second is \((x +9)(x -2)\), denominator of second is \((x +9)(x +4)\). So:

Numerator: \((x -7)(x +4) \times (x +9)(x -2)\)
Denominator: \((x +9)(x -7) \times (x +9)(x +4)\)

Cancel \((x -7)\) (top and bottom), \((x +4)\) (top and bottom), one \((x +9)\) (top and bottom). So remaining:

Numerator: \((x -2)\)
Denominator: \((x +9)\)

Wait, but let's check with x=0 again. Original expression:

First fraction: \(\frac{0 -0 -28}{0 +0 -63}=\frac{-28}{-63}=\frac{4}{9}\)
Second fraction: \(\frac{0 +0 -18}{0 +0 +36}=\frac{-18}{36}=-\frac{1}{2}\)
Multiply: \(\frac{4}{9} \times (-\frac{1}{2})=-\frac{2}{9}\)

\(\frac{x -2}{x +9}\) at x=0: \(\frac{-2}{9}\), which matches. So that's correct. Wait, but let's check another value, x=1:

Original expression:

First fraction: \(\frac{1 -3 -28}{1 +2 -63}=\frac{-30}{-60}=\frac{1}{2}\)
Second fraction: \(\frac{1 +7 -18}{1 +13 +36}=\frac{-10}{50}=-\frac{1}{5}\)
Multiply: \(\frac{1}{2} \times (-\frac{1}{5})=-\frac{1}{10}\)

\(\frac{x -2}{x +9}\) at x=1: \(\frac{1 -2}{1 +9}=\frac{-1}{10}\), which matches. So correct.

Wait, but earlier I thought maybe I messed up, but no—so the simplified form is \(\frac{x - 2}{x + 9}\)? Wait, no, wait: wait, the original problem says "leave your answer in factored form". Wait, but \(\frac{x - 2}{x + 9}\) is already simplified, and (x - 2) and (x + 9) are factored. Wait, but let's check the factoring again. Wait, maybe I made a mistake in the factoring of \(x^2 +2x -63\). Wait, no: \(x^2 +2x -63=(x +9)(x -7)\) (correct). \(x^2 -3x -28=(x -7)(x +4)\) (correct). \(x^2 +7x -18=(x +9)(x -2)\) (correct). \(x^2 +13x +36=(x +9)(x +4)\) (correct). So when we multiply:

\[ \frac{(x -7)(x +4)}{(x +9)(x -7)} \times \frac{(x +9)(x -2)}{(x +9)(x +4)} = \frac{(x -7)(x +4)(x +9)(x -2)}{(x +9)(x -7)(x +9)(x +4)} \]

Cancel (x -7), (x +4), and one (x +9):

So numerator: (x -2)
Denominator: (x +9)

Wait, but that's a linear over linear, which is factored (since they are binomials). So the simplified factored form is \(\frac{x - 2}{x + 9}\). Wait, but let me check once more. Maybe I missed a factor. Wait, no—all quadratics are factored into linear terms, then we canceled the common linear factors. So the result is \(\frac{x - 2}{x + 9}\).

Answer:

\(\frac{x - 2}{x + 9}\)