To find the greatest common factor, we must break each term into its prime factors:

\(\displaystyle \small x^2y^3z^2 = x \cdot x \cdot y \cdot y \cdot y \cdot z \cdot z\)

\(\displaystyle \small x^4y^3z= x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot z\)

The terms have \(\displaystyle \small x^2\), \(\displaystyle \small y^3\), and \(\displaystyle \small z\) in common; thus, the GCF is \(\displaystyle \small x^2y^3z\).

Pull this out of the expression to find the answer: \(\displaystyle \small x^2y^3z(z+x^2)\). 

Consider the possible values for (x, y):

(1, 100)

(2, 50)

(4, 25)

(5, 20)

Note that (10, 10) is not possible since the two variables must be distinct. The sums of the above pairs, respectively, are:

1 + 100 = 101

2 + 50 = 52

4 + 25 = 29

5 + 20 = 25, which is the smallest sum and therefore the correct answer.

\(\displaystyle y-7=\frac{4x+6y}{3}\) 

Multiply both sides by 3:

\(\displaystyle \frac{3}{1}(y-7)=(\frac{4x+6y}{3})\frac{3}{1}\) 

\(\displaystyle 3(y-7)={4x+6y}\) 

Distribute:

\(\displaystyle 3y-21={4x+6y}\)

Subtract \(\displaystyle 3y\) from both sides:

\(\displaystyle -21=4x+6y-3y\)

Add the \(\displaystyle y\) terms together, and subtract \(\displaystyle 4x\) from both sides:

\(\displaystyle -4x-21=3y\)

Divide both sides by \(\displaystyle 3\):

\(\displaystyle \frac{-4x-21=3y}{3}\)

Simplify:

\(\displaystyle y=-\frac{4}{3}x-7\)

Let's split the four terms into two groups, and find the GCF of each group.

First group: \(\displaystyle 4ab-6b\)

Second group: \(\displaystyle 28a-42\)

The GCF of the first group is \(\displaystyle 2b\).  When we divide the first group's terms by \(\displaystyle 2b\), we get: \(\displaystyle 2b(2a-3)\).

The GCF of the second group is \(\displaystyle 14\).  When we divide the second group's terms by \(\displaystyle 14\), we get: \(\displaystyle 14(2a-3)\).

We can rewrite the original expression,

\(\displaystyle 4ab-6b+28a-42\)

as,

\(\displaystyle 2b(2a-3)+14(2a-3)\)

The common factor for BOTH of these terms is \(\displaystyle (2a-3)\).

Dividing both sides by \(\displaystyle (2a-3)\) gives us:

\(\displaystyle (2b+14)(2a-3)\)

Let's separate the four terms of the polynomial expression into two groups, and then find the GCF (greatest common factor) for each group.

First group: \(\displaystyle ab-2a\)

Second group: \(\displaystyle b-2\)

The GCF of the first group is \(\displaystyle a\); it's the only factor both terms have in common.  Factoring the first group by its GCF gives us:

\(\displaystyle a(b-2)\)

The second group is a bit tricky.  It looks like they have no factor in common.  But, each of the terms can be divided by \(\displaystyle 1\)!  So, the GCF is \(\displaystyle 1\).

Factoring the second group by its GCF gives us:

\(\displaystyle 1(b-2)\)

We can rewrite the original expression:

\(\displaystyle ab-2a+b-2\) is the same as:

\(\displaystyle a(b-2)+1(b-2)\),

which is the same as:

\(\displaystyle (a+1)(b-2)\)

Separate the four terms into two groups, and then find the GCF of each group.

First group: \(\displaystyle 2ac+6ad\)

Second group: \(\displaystyle bc+3bd\)

The GCF of the first group is \(\displaystyle 2a\).  Factoring out \(\displaystyle 2a\) from the terms in the first group gives us:

\(\displaystyle 2a(c+3d)\)

The GCF of the second group is \(\displaystyle b\).  Factoring out \(\displaystyle b\) from the terms in the second group gives us:

\(\displaystyle b(c+3d)\)

We can rewrite the original expression,

\(\displaystyle 2ac+6ad+bc+3bd\)

as,

\(\displaystyle 2a(c+3d)+b(c+3d)\)

We can factor this as:

\(\displaystyle (2a+b)(c+3d)\)

Start by separating the four terms into two groups, and find the GCF (greatest common factor) of each group.

First group: \(\displaystyle -2ba-av\)

Second group: \(\displaystyle -2b-2v\)

The GCF of the first group is \(\displaystyle -2a\).  By factoring out \(\displaystyle -2a\) from each term in the first group, we are left with:

\(\displaystyle -2a(b+v)\)

(Remember, when dividing by a negative, the original number changes its sign!)

The GCF of the second group is \(\displaystyle -2\).  By factoring out \(\displaystyle -2\) from each term in the second group, we get:

\(\displaystyle -2(b+v)\)

We can rewrite the original expression,

\(\displaystyle -2ab-2av-2b-2v\)

as,

\(\displaystyle -2a(b+v)+-2(b+v)\)

The GCF of each of these terms is...

\(\displaystyle (b+v)\),

...so, the expression, when factored, is:

\(\displaystyle (-2a-2)(b+v)\)

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.,\(\displaystyle 5-7=-2\)).