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.
Herb K.
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
06/22/13