Question

simplify the expression.
\\(\frac{x^2 - 6x + 9}{2x^3 + 5x} \div \frac{x^2 - 2x - 3}{4x^2 + 10}\\)
a \\(\frac{x + 1}{2x}\\)
b \\(\frac{2}{x}\\)
c \\(\frac{2x - 6}{x^2 + x}\\)
d \\(\frac{x - 3}{x + 1}\\)

Explanation:

Step1: Rewrite division as multiplication

To divide by a fraction, multiply by its reciprocal. So, \(\frac{x^2 - 6x + 9}{2x^3 + 5x} \div \frac{x^2 - 2x - 3}{4x^2 + 10}=\frac{x^2 - 6x + 9}{2x^3 + 5x}\times\frac{4x^2 + 10}{x^2 - 2x - 3}\)

Step2: Factor each expression

• Factor \(x^2 - 6x + 9\): It is a perfect square trinomial, \(x^2 - 6x + 9=(x - 3)^2\)
• Factor \(2x^3 + 5x\): Factor out \(x\), \(2x^3 + 5x=x(2x^2 + 5)\)
• Factor \(4x^2 + 10\): Factor out \(2\), \(4x^2 + 10 = 2(2x^2+5)\)
• Factor \(x^2 - 2x - 3\): Find two numbers that multiply to \(- 3\) and add to \(-2\), which are \(-3\) and \(1\). So, \(x^2 - 2x - 3=(x - 3)(x+1)\)

Step3: Substitute factored forms

Substitute the factored forms into the expression: \(\frac{(x - 3)^2}{x(2x^2 + 5)}\times\frac{2(2x^2 + 5)}{(x - 3)(x + 1)}\)

Step4: Cancel common factors

• Cancel \((x - 3)\) from the numerator and denominator: \(\frac{(x - 3)\cancel{(x - 3)}}{x(2x^2 + 5)}\times\frac{2(2x^2 + 5)}{\cancel{(x - 3)}(x + 1)}\)
• Cancel \((2x^2+5)\) from the numerator and denominator: \(\frac{(x - 3)}{x\cancel{(2x^2 + 5)}}\times\frac{2\cancel{(2x^2 + 5)}}{(x + 1)}\)

Step5: Multiply the remaining factors

Multiply the numerators and denominators: \(\frac{2(x - 3)}{x(x + 1)}\)? Wait, no, wait. Wait, after canceling, we have \(\frac{(x - 3)\times2}{x\times(x + 1)}\)? Wait, no, let's re - check. Wait, the numerator after canceling is \(2(x - 3)\) and the denominator is \(x(x + 1)\)? Wait, no, wait:

Wait, \(\frac{(x - 3)^2}{x(2x^2 + 5)}\times\frac{2(2x^2 + 5)}{(x - 3)(x + 1)}=\frac{(x - 3)^2\times2(2x^2 + 5)}{x(2x^2 + 5)\times(x - 3)(x + 1)}\)

Cancel \((x - 3)\) (one from the square) and \((2x^2 + 5)\):

We get \(\frac{(x - 3)\times2}{x(x + 1)}\)? No, wait, no. Wait, \((x - 3)^2\div(x - 3)=(x - 3)\), \(2(2x^2 + 5)\div(2x^2 + 5) = 2\). So the expression becomes \(\frac{2(x - 3)}{x(x + 1)}\)? Wait, that's not one of the options. Wait, I must have made a mistake. Wait, let's re - do the factoring and canceling.

Wait, \(x^2-6x + 9=(x - 3)^2\), \(2x^3+5x=x(2x^2 + 5)\), \(4x^2 + 10=2(2x^2 + 5)\), \(x^2-2x - 3=(x - 3)(x + 1)\)

So the expression is \(\frac{(x - 3)^2}{x(2x^2 + 5)}\times\frac{2(2x^2 + 5)}{(x - 3)(x + 1)}\)

Cancel \((x - 3)\) (from \((x - 3)^2\) and \((x - 3)\)): \(\frac{(x - 3)\cancel{(x - 3)}}{x(2x^2 + 5)}\times\frac{2(2x^2 + 5)}{\cancel{(x - 3)}(x + 1)}\)

Cancel \((2x^2 + 5)\): \(\frac{(x - 3)}{x\cancel{(2x^2 + 5)}}\times\frac{2\cancel{(2x^2 + 5)}}{(x + 1)}\)

Now, multiply the numerators: \(2(x - 3)\) and denominators: \(x(x + 1)\)? Wait, but that's not matching the options. Wait, maybe I made a mistake in factoring. Wait, no, let's check the options again. Wait, option C is \(\frac{2x - 6}{x^2+x}\). Let's simplify \(\frac{2(x - 3)}{x(x + 1)}=\frac{2x-6}{x^2 + x}\) (since \(2(x - 3)=2x - 6\) and \(x(x + 1)=x^2+x\))

Answer:

C. \(\frac{2x - 6}{x^2+x}\)