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

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

To factor a polynomial of the form \(\displaystyle ax^2+bx+c\), we want to look at the factors of \(\displaystyle a\) and the factors of \(\displaystyle c\). We want to find the combination of factors which when multiplied and added together give the value of \(\displaystyle b\).

In our case, \(\displaystyle a=1\), \(\displaystyle b=-4\), and \(\displaystyle c=-5\).

The factors for \(\displaystyle a\) are \(\displaystyle (1,1)\).

The factors for \(\displaystyle c\) are\(\displaystyle (-5,1), (5,-1)\). 

Since \(\displaystyle b\) is \(\displaystyle -4\) we will want to use the factors \(\displaystyle (-5,1)\) because \(\displaystyle -5+1=-4\).

Therefore when we put these factors into the binomal form we get,

\(\displaystyle (x-5)(x+1)\).

Also see that

\(\displaystyle (x-5)(x+1)\) 

will foil out into the original polynomial, as \(\displaystyle -5+1=-4\), the coefficient for our \(\displaystyle x\) term, and \(\displaystyle -5*1=-5\), the constant.

There are two steps to this question that can be completed in either order.

1. Simplify the expression

2. Find the reciprocal of the expression

The reciprocal is equivalent to 1 divided by the number (1/x), essentially flipping the position of the numerator and denominator. Simplify first by recognizing you can factor the numerator:

\(\displaystyle \frac{ (x^2+2x-3)}{(x+3)}= \frac{(x+3)(x-1))}{(x+3)}=(x-1)\)

You can simplify the expression by cancelling out \(\displaystyle (x+3)\) on the top and bottom.

This leaves us with \(\displaystyle (x-1)\).

Now we find the reciprocal by dividing 1 by the number:

\(\displaystyle \large \frac{1}{x-1}\)

In the reverse order it would look like this:

\(\displaystyle \frac{(x+3)}{(x^2+2x-3)}=\frac{(x+3)}{(x+3)(x-1)}=\frac{1}{x-1}\)

*Always remember what the question is asking.

The key to factoring the polynomial \(\displaystyle 9x^2-y^2\) is recognizing a "product of a sum and a difference". The resulting inner terms cancel each other out in the final expanded polynomial.

\(\displaystyle (a+b)(a-b)= a^2{\color{Cyan} +ab}{\color{Magenta} -ab}-b^2= a^2-b^2\).

\(\displaystyle 9 = 3^2\)

One could use the FOIL and grid methods to factor the polynomial into 

\(\displaystyle (3x+y)(3x-y)=9x{\color{Cyan} +3xy}{\color{Magenta} -3xy}-y^2=9x^2-y^2\).

Leaving \(\displaystyle (3x-y)(3x+y)\) as the correct answer.

We are going to use factoring by grouping to factor this expression.

The expression:

\(\displaystyle 6x^2-13x+6\)

has the form:

\(\displaystyle Ax^2+Bx+C\)

In factoring by grouping, we want to split that B value into two smaller values a and b so we end up with this expression:

\(\displaystyle Ax^2+ax+bx+C\)

The rules are:

\(\displaystyle a+b=B\)

and

\(\displaystyle a*b=A*C\)

For our problem A=6 B=-13 C=6

So we have:

\(\displaystyle a+b=B=-13\)

\(\displaystyle a*b=A*C=6*6=36\)

Another way of saying this is that we are looking for factors of 36 that add up to -13

The values that work are:\(\displaystyle a=-9, b=-4\)

Plugging those values in:

\(\displaystyle 6x^2-9x-4x+6\)

Now let's group and factor out from both groups:

\(\displaystyle (6x^2-9x)+(-4x+6)\)

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

Finally let's factor  out a (2x-3) from both terms: and we get our answer:

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

To factor this you need to see what two numbers multiply together to give you \(\displaystyle 9\) but add together to give you \(\displaystyle 6\). There are a number of ways to test and find the answer (such as the cross method) but you will get 

\(\displaystyle (x+3)(x+3)\)

because \(\displaystyle 3\cdot3=9\)  and  \(\displaystyle 3+3=6\)

First, we see that the second term of this trinomial is negative, while the third term is positive. This should tell us that the binomials we factor out will have negative signs, because adding two negatives equals a negative, while multiplying them equals a positive.

Next we need to determine what are the factors of the third term (in this case, \(\displaystyle 8\)). We know that \(\displaystyle 1\) & \(\displaystyle 8\) and \(\displaystyle 2\) & \(\displaystyle 4\) are the only factors of \(\displaystyle 8\). So we must figure out which ones add together to equal \(\displaystyle -6\). As we said before, we know that both will be negative, so we test them out:

\(\displaystyle -1+(-8)=-9\)

\(\displaystyle -2+(-4)=-6\) This one is correct!

So the factors must be \(\displaystyle (x-2)(x-4)\).

To double-check, trying FOILing to see if it's correct.

First step is to get the equation into a form we can factor by moving the -4 to the left side:

\(\displaystyle x^2+6x+4=-4\)

\(\displaystyle x^2+6x+8=0\)

Now to factor this into two binomials we need factors of 8 that add up to 6: 4 and 2.

So we factor the expression to:

\(\displaystyle (x+4)(x+2)=0\)

Now if either of those binomials equal zero, we have a solution.

The first binomial equals zero when x=-4, and the second binomial equals zero when x=-2.

Thus our answer is:

 \(\displaystyle x=-4,x=-2\)