Question

multiply rational expressions - db (missed questions)revy 1
simplify the rational expression.
\\(\frac{15c^2 + 5c}{5c^2 + 20c}\\)
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multiply the rational expressions. give your answer in factored form.
\\(\frac{8x^2 - 2x - 1}{9x^2 - 3x - 2} \cdot \frac{3x^2 - 14x + 8}{10x^2 - 7x + 1}\\)
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hint: to add an exponent (\\(x^n\\)), type \exponent\ or press \^\
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Explanation:

First Problem: Simplify \(\boldsymbol{\frac{15c^2 + 5c}{5c^2 + 20c}}\)

Step1: Factor numerator and denominator

Factor out the greatest common factor (GCF) from numerator and denominator.
For the numerator \(15c^2 + 5c\), the GCF of \(15c^2\) and \(5c\) is \(5c\), so \(15c^2 + 5c = 5c(3c + 1)\).
For the denominator \(5c^2 + 20c\), the GCF of \(5c^2\) and \(20c\) is \(5c\), so \(5c^2 + 20c = 5c(c + 4)\).
Now the expression becomes \(\frac{5c(3c + 1)}{5c(c + 4)}\).

Step2: Cancel common factors

Cancel the common factor \(5c\) from the numerator and the denominator (assuming \(c
eq 0\) and \(c
eq -4\) to avoid division by zero).
After canceling, we get \(\frac{3c + 1}{c + 4}\).

Step1: Factor each quadratic expression

• Factor \(8x^2 - 2x - 1\):

We need two numbers that multiply to \(8 \times (-1) = -8\) and add to \(-2\). The numbers are \(-4\) and \(2\).
Rewrite the middle term: \(8x^2 - 4x + 2x - 1\).
Group and factor: \(4x(2x - 1) + 1(2x - 1) = (4x + 1)(2x - 1)\).

• Factor \(9x^2 - 3x - 2\):

Multiply \(9 \times (-2) = -18\). Find two numbers that multiply to \(-18\) and add to \(-3\): \(-6\) and \(3\).
Rewrite: \(9x^2 - 6x + 3x - 2\).
Group and factor: \(3x(3x - 2) + 1(3x - 2) = (3x + 1)(3x - 2)\).

• Factor \(3x^2 - 14x + 8\):

Multiply \(3 \times 8 = 24\). Find two numbers that multiply to \(24\) and add to \(-14\): \(-12\) and \(-2\).
Rewrite: \(3x^2 - 12x - 2x + 8\).
Group and factor: \(3x(x - 4) - 2(x - 4) = (3x - 2)(x - 4)\).

• Factor \(10x^2 - 7x + 1\):

Multiply \(10 \times 1 = 10\). Find two numbers that multiply to \(10\) and add to \(-7\): \(-5\) and \(-2\).
Rewrite: \(10x^2 - 5x - 2x + 1\).
Group and factor: \(5x(2x - 1) - 1(2x - 1) = (5x - 1)(2x - 1)\).

Step2: Substitute factored forms into the expression

Substitute the factored quadratics into the original expression:
\[
\frac{(4x + 1)(2x - 1)}{(3x + 1)(3x - 2)} \cdot \frac{(3x - 2)(x - 4)}{(5x - 1)(2x - 1)}
\]

Step3: Cancel common factors

Cancel \((2x - 1)\) from numerator and denominator, and cancel \((3x - 2)\) from numerator and denominator:
\[
\frac{(4x + 1)\cancel{(2x - 1)}}{(3x + 1)\cancel{(3x - 2)}} \cdot \frac{\cancel{(3x - 2)}(x - 4)}{(5x - 1)\cancel{(2x - 1)}} = \frac{(4x + 1)(x - 4)}{(3x + 1)(5x - 1)}
\]

Answer:

\(\frac{3c + 1}{c + 4}\)

Second Problem: Multiply \(\boldsymbol{\frac{8x^2 - 2x - 1}{9x^2 - 3x - 2} \cdot \frac{3x^2 - 14x + 8}{10x^2 - 7x + 1}}\) (in factored form)