Let's find the GCF of the coefficients, first.  Our coefficients are 3, 9, and -15.  The GCF is 3.  3 is the biggest number that can divide EACH of the numbers without leaving a remainder.

So, 3 is the first part of our GCF.

Next, look at the variables.  THERE IS NO COMMON VARIABLE!  The first term has an "r" variable, but the second term only has a "q" variable.  They do not have anything in common, so we cannon factor out any variables.

So, the only factor all of the terms have in common is "3," and when you divide each of the terms by 3, we get:

\(\displaystyle 3(r+3q-5q^2r^2)\)

At first, it looks like none of the the terms have any factors in common.  But here's how we factor by "grouping":  we separate the four terms into two groups, with two terms in each group.  Then, we find the GCF of each separate group.

First group: 6a + 3

Second group: 8ab + 4b

Start with the first group.  The GCF is "3," since that's the biggest number that can divide each term without leaving a remainder.  So, group 1 is:

\(\displaystyle 3(2a+1)\)

Second group.  The GCF is 4b.  Check it, if you're not sure.  That means that when group 2 is factored, we get:

\(\displaystyle 4b(2a+1)\)

So, since both group 1 and group 2 were being ADDED to each other in our original expression, we can simply add the two factored groups to each other:

\(\displaystyle 3(2a+1)+4b(2a+1)\)

The expression is still the same, we've just rewritten it in factored form.  If you multiply through the parentheses, you'll see that we still have the same original expression:

\(\displaystyle 6a+3+8ab+4b\)

But now, notice that the two factored terms have something in common!  They each have the factor

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

So, we can divide (2a+1) from both terms, leaving us with:

\(\displaystyle (2a+1)(4b+3)\)

Group the first two terms together:

\(\displaystyle 6az-2a\)

Find the GCF:

\(\displaystyle 2a(3z-1)\)

Group the second two terms together:

\(\displaystyle 12z-4\)

Find the GCF:

\(\displaystyle 4(3z-1)\)

The two factored terms both have \(\displaystyle (3z-1)\) as their GCF.

So, factoring out \(\displaystyle (3z-1)\) from each term gives us:

\(\displaystyle (2a+4)(3z-1)\)

The polynomial factors into (x + 3) (3x - 2).

3x² + 7x – 6 = (a + b)(c + d)

There must be a 3x term to get a 3x² term.

3x² + 7x – 6 = (3x + b)(x + d)

The other two numbers must multiply to –6 and add to +7 when one is multiplied by 3.

b * d = –6 and 3d + b = 7

b = –2 and d = 3

3x² + 7x – 6 = (3x – 2)(x + 3)

(x + 3) is the correct answer.

The product of the last two numbers should be 6, while the sum of the products of the inner and outer numbers should be 5x. Factors of six include 1 and 6, and 2 and 3. In this case, our sum is five so the correct choices are 2 and 3. Then, our factored expression is (x + 2)(x + 3). You can check your answer by using FOIL.

y = x² + 5x + 6

2 * 3 = 6 and 2 + 3 = 5

(x + 2)(x + 3) = x² + 5x + 6

Because both terms are perfect squares, this is a difference of squares:

\(\displaystyle \small x^2-25=x^2-5^2\)

The difference of squares formula is \(\displaystyle a^2-b^2 = (a+b)(a-b)\).

Here, a = x and b = 5.  Therefore the answer is \(\displaystyle \small (x-5)(x+5)\).

You can double check the answer using the FOIL method:

\(\displaystyle \small x^2-5x+5x-25= x^2-25\)

The solutions indicate that the answer is:

\(\displaystyle \small (x\ 1)(x\ 4)\) and we need to insert the correct addition or subtraction signs. Because the last term in the problem is positive (+4), both signs have to be plus signs or both signs have to be minus signs. Because the second term (-5x) is negative, we can conclude that both have to be minus signs leaving us with:

\(\displaystyle \small (x-1)(x-4)\)

Using the FOIL rule, only \(\displaystyle (3x+4)(x-2)\) yields the same polynomial as given in the question.

When asked to factor a difference of squares, the solution will always be the square roots of the coefficients with opposite signs in each pair of parentheses.

Here \(\displaystyle a=6,b=1,and\ c=-15\). Multiply \(\displaystyle a\) and \(\displaystyle c\) and you get \(\displaystyle -90\) which can be factored as \(\displaystyle 10\) and \(\displaystyle -9\) and when one adds \(\displaystyle 10\) and \(\displaystyle -9\) you get \(\displaystyle +1\).  Hence the quadratic equation can be rewritten as

\(\displaystyle 6x^{2} + 10x - 9x -15\)

Now you factor by grouping the first two terms and the last two terms giving us

\(\displaystyle 2x(3x+5)-3(3x+5)\)  which can be further factored resulting in

\(\displaystyle (2x - 3)(3x + 5)\)

By setting each of the two factors to 0 we get

\(\displaystyle x = \frac{3}{2} , \frac{-5}{3}\)