Factoring Quadratics with x2 > 1

[Gringo123]

I know how to factor a quadratic where the x² term is just x²

for example:
to factor x² + 5x + 6 I need to find 2 number that add to give 5 and multiply to give 6 (2 & 3). So the answer is (x + 2) (x + 3)

However, I'm getting stuck with this one:
2x² - x - 3

It seems that a different method is required to factor quadratics in which the x² value is greater than just x².
I know from trial and error that the answer is (2x - 3) (x + 1) but what is the correct procedure for working that out?

 

[Millacol88]

I know how to factor a quadratic where the x²
However, I'm getting stuck with this one:
2x² - x - 3

It seems that a different method is required to factor quadratics in which the x² value is greater than just x².
I know from trial and error that the answer is (2x - 3) (x + 1) but what is the correct procedure for working that out?

Set up your expression like this when you first start the problem:
(2x )(x )
2x * x = 2x², so that's your value of ax² taken care of. Now find two numbers that multiply to give you an integer value of -3. The other stipulation is that when you multiply the integer values from both sets of brackets by the x-values in the other bracket and add them, they must give you -x. Use up all the possible values of ax * x = ax² as you can, as well as all the combinations of numbers that multiply to give your integer value in expanded form, in this case, -3. I set it up like this:

(2x - 3)(x + 1)
We know that 2x * x = 2x², and that -3 * 1 is -3, now we have to check to make sure that when we multiply to get our x-term, it ends up being what it is in expanded form:
-3(x) + 1(2x) = -x
-3x +2x = -x
-x = -x

All you are really doing is trying different terms that multiply to give you the correct value of ax², as well as trying different integer-values that multiply to give you the integer present in expanded form. Then, check to see if, when the distributive property is used on your factored expression, whether or not your x-term matches the one in expanded form. If it does you're done.

 

[The Chaz]

...
All you are really doing is trying different terms that multiply to give you the correct value of ax², as well as trying different integer-values that multiply to give you the integer present in expanded form. Then, check to see if, when the distributive property is used on your factored expression, whether or not your x-term matches the one in expanded form. If it does you're done.

"Guess-and-check" is my least favorite way to do this, for reasons that I'll get into later.

With a quadratic polynomial in STANDARD FORM ax² + bx + c = 0
write this:
(1/a)(ax + p)(ax + q), where p + q = b and pq = ac. I know that's a lot of letters/variables, but you're used to this method (i.e. find two numbers that add-up to the middle coefficient, and whose product is ac). After you find and fill-in p and q, simplify.

Consider this:
20x² +39x + 18. Guess-and-check? NO WAY!
ac = (20)*(18) = 360. Find two numbers that multiply to give you 360, and that add up to 39. For this step, let's write factor pairs of 360, with their sums off to the side.
1*360 ...361
2*180...182
3*120...123
4*90...94 (notice that their sums get smaller. We're looking for 39)
5*72 ...77
6*60...66
7*(doesn't go evenly. I'll skip the rest like this)
8*45...53
9*40...49 (not 39, though!)
10*36...46
12*30...42
15*24...39 (BINGO)
18*20...38
20*18... but we're going back up the chain, so we quit.

Just to recap, we are looking for two numbers that add up to 39 and multiply to get 360. I call those numbers "p" and "q". We have found them. p = 15 and q = 24. (You can switch the order and it won't matter in the end)

We plug these into the equation (1/a)(ax + p)(ax + q). Don't forget that "a" is the number in front of x² in our original problem (20x² +39x + 18). So a = 20.

(1/20)(20x + 15)(20x + 24). You can factor a "5" out of (20x + 15), and a "4" out of (20x + 24)...
(1/20)(5)(4x + 3)(4)(5x + 6). The (1/20)*(4)*(5) just gives you "1"...

20x² +39x + 18 = (4x + 3)(5x + 6).

.......
Now we get into your specific question. There are some negatives, so that will provide much-needed variety.
2x² - x - 3. a = 2; b = -1; c = -3.
We want two numbers that have the sum of -1 (b) and whose product is -6 (ac). Here are the factor pairs of -6:
1*-6... with sum -5
-1*6 ...+5
2*-3 ...-1 ("bingo")
-2*3...+1
So p = 2 and q = -3.
(1/a)(ax + p)(ax + q) becomes
(1/2)(2x + 2)(2x + -3)
(1/2)*2*(x + 1)(2x - 3)
(x + 1)(2x - 3). I guess that's right ;)

 

[Gringo123]

Thanks a lot guys. That was a great help! Do you mind just looking at this one and confirming that I have the right answer?
15xsquared + 19x + 6
factored out:
(3x + 2) (5x + 3)
therefore value of x = 2/3 or 3/5
Thanks again for your help!

 

[iRaid]

ax²+bx+c
=
(#+#)(#+#)

Using reverse FOIL (First outer inner last) to get the actual numbers.. Then using normal FOIL, check to see if you're correct.

 

[Mark44]

Thanks a lot guys. That was a great help! Do you mind just looking at this one and confirming that I have the right answer?
15xsquared + 19x + 6
factored out:
(3x + 2) (5x + 3)
therefore value of x = 2/3 or 3/5
Thanks again for your help!

You should get in the habit of checking your work yourself. If (3x + 2)(5x + 3) multiplies to make 15x² + 19x + 6, then you have factored the trinomial correctly. You've already done the hard work of factoring the trinomial into two binomial factors. Checking is easy - just multiply the two binomial factors.

For the other part of your post here, those answers are wrong possibly for two reasons.
1. If all you are doing is factoring 15x² + 19x + 6, you are just rewriting this expression in a different form, a factored form. Once you have done that, you are done. 15x² + 19x + 6 is not an equation, so it is not possible to find solutions. If you know that 15x² + 19x + 6 = 0, you now have an equation, so you can find the solutions by factoring or using the quadratic formula.
2. If the goal was to find the solution to the equation 15x² + 19x + 6 = 0 (which isn't evident from what you have posted), and if this factorization is correct (I'm not saying) (3x + 2)(5x + 3) = 0, then the solutions are NOT x = 2/3 and x = 3/5.

 

[The Chaz]

Thanks a lot guys. That was a great help! Do you mind just looking at this one and confirming that I have the right answer?
15xsquared + 19x + 6
factored out:
(3x + 2) (5x + 3)
therefore value of x = 2/3 or 3/5
Thanks again for your help!

I don't check; I just make sure I do it right at each step! (I know that sounds pompous, but at some point you need confidence...and this method is foolproof IF you factor out the GCF first)

How did you get your values of x? Did you solve the equation 3x+ 2= 0?
(Hint: no, you didn't! Close...)

 

[Gringo123]

I know that (3x + 2) (5x + 3) = 0
if 3x + 2 = 0, then 3x = 2. Dividing both sides by 3 gives x = 2/3
like wise if 5x = 3, dividing both sides by 5 gives x = 3/5

 

[The Chaz]

I know that (3x + 2) (5x + 3) = 0
if 3x + 2 = 0, then 3x = 2. (There is no justification for this step. i.e. "it's wrong")

Dividing both sides by 3 gives x = 2/3
like wise if 5x = 3, dividing both sides by 5 gives x = 3/5

3x + 2 = 0 is an equality, so we must do the same thing to both sides (subtract 2)
3x = -2
divide by 3
x = -2/3

The other gives x = -3/5.

 

[Gringo123]

Thanks a lot guys!