Rewrite this as \(\displaystyle 3y^{2} + 1 y - 10\)

Use the \(\displaystyle ac\)-method by splitting the middle term into two terms, finding two integers whose sum is 1 and whose product is \(\displaystyle 3 (-10) = -30\); these integers are \(\displaystyle -5,6\), so rewrite this trinomial as follows:

\(\displaystyle 3y^{2} + 1 y - 10 = 3y^{2} -5y + 6 y - 10\)

Now, use grouping to factor this:

\(\displaystyle \left (3y^{2} -5y \right )+ \left (6 y - 10 \right )\)

\(\displaystyle = y \left (3y -5 \right )+ 2 \left (3y -5 \right )\)

\(\displaystyle = \left (y+ 2 \right ) \left (3y -5 \right )\)

\(\displaystyle 7x^2+56+C\)

Factor out the 7:

\(\displaystyle 7\left (x^2+8x+\frac{C}{7} \right )\)

Take the 8 from the x-term, cut it in half to get 4, then square it to get 16.  Make this 16 equal to C/7:

\(\displaystyle 16=\frac{C}{7}\)

Solve for C:

\(\displaystyle C=112\)

We can factor this trinomial using the FOIL method backwards. This method allows us to immediately infer that our answer will be two binomials, one of which begins with \(\displaystyle 3x\) and the other of which begins with \(\displaystyle x\). This is the only way the binomials will multiply to give us \(\displaystyle 3x^{2}\).

The next part, however, is slightly more difficult. The last part of the trinomial is \(\displaystyle -2\), which could only happen through the multiplication of 1 and 2; since the 2 is negative, the binomials must also have opposite signs.

Finally, we look at the trinomial's middle term. For the final product to be \(\displaystyle -x\), the 1 must be multiplied with the \(\displaystyle 3x\) and be negative, and the 2 must be multiplied with the \(\displaystyle x\) and be positive. This would give us \(\displaystyle \left ( -3x\right )+\left ( 2x\right )\), or the \(\displaystyle -x\) that we are looking for.

In other words, our answer must be 

\(\displaystyle \left ( 3x+2\right )\left ( x-1\right )\) 

to properly multiply out to the trinomial given in this question.

One way to factor a trinomial like this one is to put the terms of the polynomial into a box/grid:

Notice that there are 4 boxes but only 3 terms. To fix this, we find two numbers that add to -7 [the middle coefficient] and multiply to -30 [the product of the first and last coefficients]. By examining the factors of -30, we discover that these numbers must be -10 and 3.

Now we can put the terms into the box, by splitting the -7x into -10x and 3x:

To finish factoring, determine the greatest common factor of each of the rows and columns. For instance, \(\displaystyle 2x^2\) and \(\displaystyle -10x\) have a greatest common factor of \(\displaystyle 2x\), and \(\displaystyle 3x\) and -15 have a greatest common factor of 3.

Our final answer just combines the factors on the top and side into binomials. In this case, \(\displaystyle (x-5)(2x+3)\).

The first step in setting up our binomial is to figure out our possible leading terms. In this case, the only reasonable factors of \(\displaystyle x^2\) are \(\displaystyle x \cdot x\):

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

Next, find the factors of \(\displaystyle 6\). In this case, we could have \(\displaystyle 6 \cdot 1\) or \(\displaystyle 3 \cdot 2\). Either combination can potentially produce \(\displaystyle 5x\), so the signage is important here.

Note that since the last term in the ordered trinomial is positive, both factors must have the same sign. Further, since the middle term in the ordered triniomial is negative, we know the signs must both be negative.

Therefore, we have two possibilities, \(\displaystyle (x-1)(x-6)\) or \(\displaystyle (x-3)(x-2)\).

Let's solve for both, and check against the original triniomial.

\(\displaystyle (x-1)(x-6) = (x^2 -x - 6x + 6) = (x^2 - 7x +6) \neq x^2 -5x + 6\)

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

Thus, our factors are \(\displaystyle (x-3)\) and \(\displaystyle (x-2)\).

The first step in setting up our binomial is to figure out our possible leading terms. In this case, the only reasonable factors of \(\displaystyle x^2\) are \(\displaystyle x \cdot x\):

\(\displaystyle x^2-2x - 8= (x + \cdots)(x+ \cdots)\)

Next, find the factors of \(\displaystyle 8\). In this case, we could have \(\displaystyle 8 \cdot 1\) or \(\displaystyle 4 \cdot 2\). Since the factor on our leading term is \(\displaystyle 1\), and no additive combination of \(\displaystyle 8\) and \(\displaystyle 1\) can create \(\displaystyle 2\), we know that our factors must be \(\displaystyle 4 \cdot 2\).

Note that since the last term in the ordered trinomial is negative, the factors must have different signs.

Therefore, we have two possibilities, \(\displaystyle (x-4)(x+2)\) or \(\displaystyle (x+4)(x-2)\).

Let's solve for both, and check against the original triniomial.

\(\displaystyle (x-4)(x+2) = (x^2 -4x + 2x -8) = (x^2 -2x -8)\)

\(\displaystyle (x+4)(x-2) = (x^2 +4x -2x -8) = (x^2 +2x - 8) \neq (x^2 -2x -8)\)

Thus, our factors are \(\displaystyle (x-4)\) and \(\displaystyle (x+2)\).

The first step in setting up our binomial is to figure out our possible leading terms. In this case, the only reasonable factors of \(\displaystyle x^2\) are \(\displaystyle x \cdot x\):

\(\displaystyle x^2+8x+15= (x + \cdots)(x+ \cdots)\)

Next, find the factors of \(\displaystyle 15\). In this case, we could have \(\displaystyle 1 \cdot 15\) or \(\displaystyle 3 \cdot 5\). Since the factor on our leading term is \(\displaystyle 1\), and no additive combination of \(\displaystyle 1\) and \(\displaystyle 15\) can create \(\displaystyle 8\), we know that our factors must be \(\displaystyle 3 \cdot 5\).

Note that since the last term in the ordered trinomial is positive, the factors must have different signs. Since the middle term is also positive, the signs must both be positive.

Therefore, we have only one possibility, \(\displaystyle (x+3)(x+5)\).

Let's solve, and check against the original triniomial.

\(\displaystyle (x+3)(x+5) = (x^2 +3x + 5x +15) = (x^2+8x+15)\)

Thus, our factors are \(\displaystyle (x+3)\) and \(\displaystyle (x+5)\).

The first step in setting up our binomial is to figure out our possible leading terms. In this case, the only reasonable factors of \(\displaystyle x^2\) are \(\displaystyle x \cdot x\):

\(\displaystyle x^2-11x+28= (x + \cdots)(x+ \cdots)\)

Next, find the factors of \(\displaystyle 24\). In this case, we could have \(\displaystyle 1, 2, 4, 7, 14,\) or \(\displaystyle 28\). Since the factor on our leading term is \(\displaystyle 1\), and no additive combination of \(\displaystyle 28\) and \(\displaystyle 1\)  or of \(\displaystyle 14\) and \(\displaystyle 2\) can create our middle term of \(\displaystyle 11\), we know that our factors must be \(\displaystyle 4 \cdot 7\).

Note that since the last term in the ordered trinomial is positive, the factors must have the same sign. Since the middle term is negative, both factors must have a negative sign.

Therefore, we have one possibility, \(\displaystyle (x-4)(x-7)\).

Let's solve, and check against the original triniomial, before solving for \(\displaystyle x=4\).

\(\displaystyle (x-7)(x-4) = (x^2 -7x -4x +28) = (x^2 - 11x + 28)\)

Thus, our factors are \(\displaystyle (x-4)\) and \(\displaystyle (x-7)\).

Now, let's solve for \(\displaystyle x=4\). Simply plug and play:

\(\displaystyle (x-4) = (4-4) = 0\)

\(\displaystyle (x-7) = (4-7) = -3\)

Therefore, \(\displaystyle x=0\) and \(\displaystyle x=-3\).

Note that, in algebra, we can represent this by showing \(\displaystyle x = 0\vee-3\). However,  writing the word "and" is perfectly acceptable.

A factored trinomial is in the form \(\displaystyle (x + p) (x + q)\), where \(\displaystyle p + q =\) the second term and \(\displaystyle pq =\) the third term.

To factor a trinomial you first need to find the factors of the third term. In this case the third term is \(\displaystyle -10\). 

Factors of \(\displaystyle -10\) are: \(\displaystyle (1, -10), (-1, 10), (2, -5), (-2, 5)\)

The factors you choose not only must multiply to equal the third term, they must also add together to equal the second term.

In this case they must equal \(\displaystyle 3\).

\(\displaystyle (-2, 5): -2 \cdot 5 = -10\)

\(\displaystyle (-2, 5): -2 + 5 = 3\)

To check your answer substitute the factors of \(\displaystyle 10\) into the binomials and use FOIL.

\(\displaystyle x^2 + 3x -10\)

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

First terms: \(\displaystyle x \cdot x = x^2\)

Outside terms: \(\displaystyle x \cdot 5 = 5x\)

Inside terms: \(\displaystyle -2 \cdot x = -2x\)

Last terms: \(\displaystyle -2 \cdot 5 = -10\)

\(\displaystyle x^2 + 5x -2x -10\)

Simplify from here by combining like terms

\(\displaystyle x^2 + 3x -10\)

Using FOIL returned it to the original trinomial, therefore the answer is:

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