You can set the two numbers to equal variables, so that you can set up the algebra in this problem. The first odd number can be defined as $\displaystyle x$ and the second odd number, since the two numbers are consecutive, will be $\displaystyle x+2$.

This allows you to set up the following equation to include the given product of 195:

$\displaystyle x(x+2)=195$

$\displaystyle x^2 + 2x = 195$

Next you can subtract 195 to the left and set the equation equal to 0, which allows you to solve for $\displaystyle x$:

$\displaystyle x^2+2x-195=0$

You can factor this quadratic equation by determining which factors of 195 add up to 2. Keep in mind they will need to have opposite signs to result in a product of negative 195:

$\displaystyle (x \ \ \ \ \ $(x \ \ \ \ \ \) = 0\)

$\displaystyle (x -13)(x +15) = 0$

Set each binomial equal to 0 and solve for $\displaystyle x$. For the purpose of this problem, you'll only make use of the positive value for $\displaystyle x$:

$\displaystyle x-13 = 0 \ \ \ \ \ \ x +15 = 0$

$\displaystyle x = 13$

Now that you have solved for $\displaystyle x$, you know the two consecutive odd numbers are 13 and 15. You solve for the answer by finding the sum of these two numbers:

$\displaystyle 13+15=28$

The common factor here is $\displaystyle x$. Pull this out of both terms to simplify:

$\displaystyle \frac{x(x^2+3x)}{x}=x(\frac{x^2}{x}+\frac{3x}{x})=x(x+3)$

Because the $\displaystyle x^2$ term doesn’t have a coefficient, you want to begin by looking at the $\displaystyle c$ term ($\displaystyle ax^2+bx+c$) of the polynomial: $\displaystyle -35$.  Find the factors of $\displaystyle -35$ that when added together equal the second coefficient (the $\displaystyle b$ term) of the polynomial. 

There are only four factors of $\displaystyle 35$: $\displaystyle 1, 5, 7, 35$, and only two of those factors, $\displaystyle 5, 7$, can be manipulated to equal $\displaystyle -2$ when added together and manipulated to equal $\displaystyle -35$ when multiplied together: $\displaystyle 5, -7$ (i.e.,$\displaystyle 5-7=-2$). 

Because the $\displaystyle x^2$ term doesn’t have a coefficient, you want to begin by looking at the $\displaystyle c$ term ($\displaystyle ax^2+bx+c$) of the polynomial: $\displaystyle 30$. 

Find the factors of $\displaystyle 30$ that when added together equal the second coefficient (the $\displaystyle b$ term) of the polynomial: $\displaystyle 13$. 

There are seven factors of $\displaystyle 30$: $\displaystyle 1, 2, 3, 5, 6, 10, 15$, and only two of those factors, $\displaystyle 3, 10$, can be manipulated to equal $\displaystyle 13$ when added together and manipulated to equal $\displaystyle 30$ when multiplied together: $\displaystyle 3, 10$ 

$\displaystyle 3+10=13, 3\times10=30$

First, factor the numerator: $\displaystyle (x+5)(x-3)$.  

Now your expression looks like 

$\displaystyle \frac{(x+5)(x-3)}{x+5}$

Second, cancel the "like" terms - $\displaystyle (x+5)$ - which leaves us with $\displaystyle x-3=10$.  

Third, solve for $\displaystyle x$, which leaves you with $\displaystyle x=13$. 

Because the $\displaystyle x^2$ term has a coefficient, you begin by multiplying the $\displaystyle a$ and the $\displaystyle c$ terms ($\displaystyle ax^2+bx+c$) together: $\displaystyle 2\times-3=-6$. 

Find the factors of $\displaystyle -6$ that when added together equal the second coefficient (the $\displaystyle b$ term) of the polynomial: $\displaystyle -5$. 

There are four factors of $\displaystyle 6$: $\displaystyle 1, 2, 3, 6$, and only two of those factors, $\displaystyle 1, 6$, can be manipulated to equal $\displaystyle -5$ when added together and manipulated to equal $\displaystyle -6$ when multiplied together: $\displaystyle 1, -6$ 

$\displaystyle 1+-6=-5, 1\times-6=-6$

For each term in this expression, we will notice that each shares a variable of $\displaystyle x$.  This can be pulled out as a common factor.

$\displaystyle x^6+2x^3+2x = x(x^5+2x^2+2)$

There are no more common factors, and this is the reduced form.

The answer is:  $\displaystyle x(x^5+2x^2+2)$

For each term in this expression, we will notice that each shares a variable of $\displaystyle x$.  This can be pulled out as a common factor.

$\displaystyle x^6+2x^3-3x = x(x^5+2x^2-3)$

There are no more common factors, and this is the reduced form.

The answer is:  $\displaystyle x(x^5+2x^2-3)$