Factoring is essentially, simplifying the expression. Usually it is used to find the roots of an equation. When you have a number in front of the x^2, you should always attempt to factor out that number. Meaning, all the other numbers can be divided by that
number. For example: 2x^2+2x+10 could be 2(x^2+2+5). Then one could factor the expression further without the 2. However, in this case, the 4 cannot be factored out.

My next step is to determine the factors of 30.

1,30 2,15 3,30 5,6

Now that we have the factors, we can try and figure out which solution works best.

My first option is (2x+____)(2x+____). In this case, it would not matter which number went in which spot because there are identical. However through trial and error, one can see that not of our factors will finish the solution.

The second option is (4x+__)(x+___). In this case, it does matter which number goes in which spot. The best way to figure this out is through trial and error.

The answer is (4x+5)(x+6).

If we were to expand the above equation we would get:

4x^2+24x+5x+30=4x^2+29x+30. The answer works.

This is how I knew the numbers were 5 and 6 and why they went in each spot.

Again our factor option are: 1,30 2,15 3,10 and 5,6

If we put the 30,15, or 10 in the (x+___) spot, it would be multiplied by 4 and be over 30 so you know that none of those answers could go there.

(4x+30)(x+1)=4x^2+4x+30x+30=4x^2+34x+30. Our solution does not match. The "x" value is too high.

(4x+15)(x+2)=4x^2+8x+15x+30=4x^2+23x+30. Our solution does not match. The "x" value is too low.

(4x+10)(x+3)=4x^2+12x+10x+30=4x^2+22x+30. Our solution does not match. The "x" value is too low.

(4x+6)(x+5)=4x^2+20x+6x+30=4x^2+26x+30. Out solution does not match. The "x" value is too low.

This is why (4x+5)(x+6) is the correct answer.

## Comments

Note: to understand why the above method of factoring works, consider the following ---

write the general Quadratic Trinomial, in factored form, as follows:

Ax^2 + Bx + C = (ax + b)(cx + d), where A, B, and C are given (for example, A = 4; B = 29; C = 30) and where a, b, c, and d are to be determined

then, using the Distributive Property (i.e., FOIL), write the following:

Ax^2 + Bx + C = (ac)x^2 + (ad + bc)x + bd, implying A = ac; B = ad + bc; C = bd;

now, note that: AC = (ac)(bd) = (ad)(bc);

thus, we're looking for two numbers, "ad" and "bc", whose product is

(ad)(bc) = AC, and whose sum is B = ad + bc

the remainder of the argument then follows, as in the example above

one further note --- if it turns out that we can find no numbers a, b, c, d such that

(ac)(bd) = (ad)(bc) = AC, and (ad + bc) = B,

then we conclude that Ax^2 + Bx + C is prime, i.e., non-factorable (at least, non-factorable "over the integers", i.e., with each of a, b, c, d being integers).

one last note --- from the previous note, we see that one example of a non-factorable Quadratic Trinomial would be the following: 4x^2 + 31x + 30, as would any Quadratic Trinomial of the form 4x^2 + Bx + 30, where B is any of the numbers {30, 31, 32, 33, 34, 35, 36, 37, 38, etc., etc., etc., ....} --- note that if, in the above list, B is even (a multiple of 2), then a factor of 2 could be factored out from 4x^2 + Bx + 30, but apart from that, 4x^2 + Bx + 30 would be non-factorable, i.e., it would not be factorable as a product of two linear factors, each of the form (ax + b)

a related question for teachers of the above topic could be the following:

to continue the previously interrupted comment:

a related question for teachers of the above topic could be the following:

for what integer values of "B" is 4x^2 + Bx + 30 factorable?