# When to use x^2 coefficient in bracket and when to factorise outside bracket

#### Simonsky

##### Junior Member
Not sure I've worded this well but here goes.

When the x^2 has a coefficient >1 I tend to use the method where you split the linear term in two. But in these cases:

4x^2 + 19x + 12

you can't do that. So is there a general rule that when you can't do that it means one of the brackets MUST contain the x^2 coefficient?

#### Dr.Peterson

##### Elite Member
Not sure I've worded this well but here goes.

When the x^2 has a coefficient >1 I tend to use the method where you split the linear term in two. But in these cases:

4x^2 + 19x + 12

you can't do that. So is there a general rule that when you can't do that it means one of the brackets MUST contain the x^2 coefficient?
This polynomial can be factored (as I can determine quickly from its discriminant); so the method you want to use will work. Please show your work from which you concluded that it can't be done.

#### Subhotosh Khan

##### Super Moderator
Staff member
Not sure I've worded this well but here goes.

When the x^2 has a coefficient >1 I tend to use the method where you split the linear term in two. But in these cases:

4x^2 + 19x + 12

you can't do that. So is there a general rule that when you can't do that it means one of the brackets MUST contain the x^2 coefficient?
4 * 12 = 16 * 3

and

16 + 3 = 19

#### HallsofIvy

##### Elite Member
$$\displaystyle (ax+ b)(cx+ d)= acx^2+ (ad+ bc)x+ bd$$. So you want to find a, b, c, and d so that ac= 4, ad+ bc= 19, and bd= 12. 4 can be factored as 1*4 or as 2*2. 12 can be factored as 1*12, 2*6, or 3*4.

Trying a= 1, c= 4, ad+ bc= d+ 4b= 19 and I notice that with d= 3, b= 4, we have 3+ 4(3)= 19! So $$\displaystyle (x+ 4)(4x+ 3)= 4x^2+ (16+ 3)x+ 12= 4x^2+ 19x+ 12$$.

Of course, it can happen that no integer values of a, b, c, and d will work. "Almost all" polynomials of the form $$\displaystyle ax^2+ bx+ c$$ cannot be factored with integer coefficients.

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#### Simonsky

##### Junior Member
Apologies for wasting your time! I can now see it is easy to factorise using the method of splitting the linear term:

4x^2 + 19x +12

4x^2 + 16x + 3x +12

4x(x+4) =3(x+4)

(4x+3)(x+4)

Sometimes a suffer from 'brain freeze' and can't see that something is really simple! Wood for the Trees job!

#### Dr.Peterson

##### Elite Member
Apologies for wasting your time! I can now see it is easy to factorise using the method of splitting the linear term:

4x^2 + 19x +12

4x^2 + 16x + 3x +12

4x(x+4) =3(x+4)

(4x+3)(x+4)

Sometimes a suffer from 'brain freeze' and can't see that something is really simple! Wood for the Trees job!
I presume your difficulty was in not seeing the pair 16+3 as satisfying the requirements. Perhaps if you tell us how you tried to find those numbers, we could help you with a method for doing that that doesn't leave you thinking it can't be done. An important part of the factoring process is to be able to be sure whether a polynomial is prime or not.

#### Simonsky

##### Junior Member
I presume your difficulty was in not seeing the pair 16+3 as satisfying the requirements. Perhaps if you tell us how you tried to find those numbers, we could help you with a method for doing that that doesn't leave you thinking it can't be done. An important part of the factoring process is to be able to be sure whether a polynomial is prime or not.
I think what happened in my thinking was a confusion between the product and sum and its relationship to the constant-so my brain toggled between the product equaling the constant and the product equaling the product of coefficient of x^2 and the constant. Holding a concept steady is still a challenge to me!

#### Dr.Peterson

##### Elite Member
I think what happened in my thinking was a confusion between the product and sum and its relationship to the constant-so my brain toggled between the product equaling the constant and the product equaling the product of coefficient of x^2 and the constant. Holding a concept steady is still a challenge to me!
Okay, that helps. I find a lot of students have this sort of problem, either losing focus or just forgetting which is sum and which is product.

What I recommend to them is to write down what they are thinking, so they have something to focus on. I tell them, in general, "Think. Then write what you thought. Then think about what you wrote. Then fix it!"

Here, you are factoring 4x^2 + 19x + 12 using the method that requires you to split 19 into the sum of two numbers whose product is 48 (the product of 4 and 12). So we write down:

product = 48
sum = 19.

Then we think -- is that really what I want to do? Either we can go back to the reasoning behind the method (which unfortunately is not always taught); or just keep some basic fact in mind, such as that the product of our two numbers is the same as the product of a and c. That can make it a little more memorable.

Now, having the goal written down, we can carry it out. I don't know how you do your search for numbers, but I either just see a pair that works, or, if that doesn't come to me quickly, start listing pairs of factors of 48:

1*48
2*24
3*16

until I get a pair whose sum is 19, as I just did.

Whether it just came to me or I went through an orderly list, I then check my pair against the criteria I wrote down. Is the sum 19? Yes, 3 + 16 = 19. Is the product 48? Yes, 3*16 = 48. This is especially important when I might get signs wrong, or I might accidentally see a pair of numbers whose sum is 48 and whose product is 19. (For example, if the product were 11 and sum -12.)

Now, the error you specifically mentioned is forgetting whether to make the product c or ac. That is an easy thing to neglect because when a=1, you do just use c. But in that case, c is in fact the same as ac. So what you can do is get used to always using ac, regardless of what a is. So whenever you factor a quadratic trinomial, you can just recite "the product is the product", or whatever helps you.

#### Simonsky

##### Junior Member
Okay, that helps. I find a lot of students have this sort of problem, either losing focus or just forgetting which is sum and which is product.

Now, the error you specifically mentioned is forgetting whether to make the product c or ac. That is an easy thing to neglect because when a=1, you do just use c. But in that case, c is in fact the same as ac. So what you can do is get used to always using ac, regardless of what a is. So whenever you factor a quadratic trinomial, you can just recite "the product is the product", or whatever helps you.
That's very helpful Dr. Peterson. You emphasis on being aware of what is being thought is crucial. The fact that we are still using ac when the coefficient is 1 helps because then the splitting of the linear component doesn't feel like its another beast! I'm finding this maths thing a real journey into an aspect of thinking that I am not used to -it's a bit of a shock to discover that I'm incredibly sloppy but I guess it's a case of persistence and practise until something in the mind changes and sees connections and meanings. Maybe it's like a musician who has to keep playing the exercise.

#### JeffM

##### Elite Member
It seems to me that it might help to point out explicitly something alluded to in Dr. Peterson's first post.

You can factor any quadratic by using the quadratic formula. (In fact, I teach those I tutor to use the quadratic formula if they do not quickly "see" a factoring.)

You have the quadratic $$\displaystyle 4x^2 + 19x + 12.$$

The discriminant of that quadratic is $$\displaystyle 19^2 - 4 * 4 * 12 = 361 - 192 = 169 \ge 0.$$

Therefore the quadratic can be factored in the real numbers.

The square root of that discriminant is $$\displaystyle \sqrt{169} = 13 \in \mathbb Q.$$

Therefore the quadratic can be factored in the rational numbers.

The roots of that quadratic are:

$$\displaystyle \dfrac{-\ 19 \pm 13}{2 * 4} = -\ \dfrac{6}{8} = -\ \dfrac{3}{4} \text { or } -\ \dfrac{32}{8} = -\ 4.$$

$$\displaystyle \therefore 4x^2 + 19x + 12 = 4 * \left \{ x - \left (-\ \dfrac{3}{4} \right ) \right \} * \{x - (-\ 4)\} = (4x + 3)(x + 4).$$

Using the quadratic formula is not very fast, but it is quite mechanical. No stress.

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