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paulb203

Member

Registered: 2023-02-24

Posts: 136

Factorising a quadratic expression where the a term is greater than 1

Example:

2x^2 + 9x + 4  (ax^2 + bx + c)

My way of doing this would be to write out a pair of brackets, 2x in one, x in the other. Then list the factors of 4. Then choose the pair that will give 9 when one of the pair is multiplied by 2, and the other one is multiplied by 1. In this case the pair would be 4, and 1. So:

(2x +1) (x+4).

*

But I've been shown a new method which is puzzling me.

Step 1. Multiply the a term by the c term (the 2 by the 4) to give 8.  Then write out a pair of brackets but this time put 2x in both of them, like this:

(2x      ) (2x     )

Then list the factors of 8. Then choose the pair which add to give 9, in this case 1, and 8. Then put the pair into the brackets:

(2x + 1) (2x +8 )

Then see where you can cancel:

(2x +1) remains the same, but (2x + 8) can be divided by 2 to become (x+4)

So our answer is:

(2x+1)(x+4)

Here is where I'm puzzled. I can use the second method to get the correct answer, and it's quicker, I think, than the first method. But I understand the first method. I know why I am taking each step. With the second method I'm left wondering why does it work.

Last edited by paulb203 (2023-10-06 09:46:39)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,196

Re: Factorising a quadratic expression where the a term is greater than 1

I'm not seeing the method here. Suppose I have

So I start with (3x    )(3x    ) but then what?

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

amnkb

Member

Registered: 2023-09-19

Posts: 253

Re: Factorising a quadratic expression where the a term is greater than 1

Example: 2x^2 + 9x + 4

My way of doing this would be to write out a pair of brackets, 2x^2 in one, x in the other. Then list the factors of 4. Then choose the pair that will give 9 when one of the pair is multiplied by 2, and the other one is multiplied by 1. In this case the pair would be 4, and 1. So:

(2x^2 +1) (x+4)

but this doesnt multiply back to what you started with
you'll get 2x^3+8x^2+x+4

But I've been shown a new method which is puzzling me.

Step 1. Multiply the a term by the c term (the 2 by the 4) to give 8.  Then write out a pair fo brackets but his time put 2x in both of them, like this:

(2x      ) (2x     )

where did you see this?
youre right its confusing!

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Factorising a quadratic expression where the a term is greater than 1

Sorry, I made a mistake on the first example. The brackets should've contained 2x, and x, to begin with, not "2^2 and x

Example: 2x^2 + 9x + 4

My way of doing this would be to write out a pair of brackets, 2x^2 in one, x in the other. Then list the factors of 4. Then choose the pair that will give 9 when one of the pair is multiplied by 2, and the other one is multiplied by 1. In this case the pair would be 4, and 1. So:

(2x^2 +1) (x+4)

but this doesnt multiply back to what you started with
you'll get 2x^3+8x^2+x+4

But I've been shown a new method which is puzzling me.

Step 1. Multiply the a term by the c term (the 2 by the 4) to give 8.  Then write out a pair fo brackets but his time put 2x in both of them, like this:

(2x      ) (2x     )

where did you see this?
youre right its confusing!

Last edited by paulb203 (2023-10-05 09:10:46)

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Factorising a quadratic expression where the a term is greater than 1

Multiply the a term by the c term (3 x 8 = 24). Then take the pair of factors of 24 that give the b term (14), which would be 2 and 12. Then put the 2 in one bracket, and the 12 in the other. Then cancel the 3x+12 to make it x + 4. And you'll have your factors; (3x+2)(x+4), which equal 3x^2 + 14x + 8.

I'm not seeing the method here. Suppose I have

So I start with (3x    )(3x    ) but then what?

Bob

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Factorising a quadratic expression where the a term is greater than 1

@amnkb

I don't find the method confusing (see my reply to Bob), as in, how to get the correct answer. I just don't know WHY it works. You asked where I saw it. The teaching assistant in my maths class showed me it. It does save on the trial and error of the first method I showed.

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Factorising a quadratic expression where the a term is greater than 1

P.S. I've edited the mistake in the first example in my original post, so's not to confuse any newcomers to the thread!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,196

Re: Factorising a quadratic expression where the a term is greater than 1

It seems to me that you still have to experiment with factors to make the middle term.

I've got to find two suitable factors of 8. Could be 1 and 8; could be 2 and 4.  I have to notice that 3x4 + 2x1 = 14.  That's no different from what you usually have to do.

I think this is just a slightly different twist on trying to 'fiddle' the middle term by experimenting with factors for the constant term. What I'd usually do is try 3 and 1 with 1 and 8; and try 3 and 1 with 2 and 4.  I'd get to 3x4 + 2x1 just the same.

Can you show me an example where it makes a real difference?

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Factorising a quadratic expression where the a term is greater than 1

Thanks, Bob.

“It seems to me that you still have to experiment with factors to make the middle term.
I've got to find two suitable factors of 8. Could be 1 and 8; could be 2 and 4.  I have to notice that 3x4 + 2x1 = 14.  That's no different from what you usually have to do.”

We’re no longer looking for factors of 8, at least not initially. Once we multiply the a term (3) by the c term (8) we get 24.
Then we list the factors of 24. From those we can immediately see that 2 and 12 add to give the b term, the middle term (14).

“I think this is just a slightly different twist on trying to 'fiddle' the middle term by experimenting with factors for the constant term. What I'd usually do is try 3 and 1 with 1 and 8; and try 3 and 1 with 2 and 4.  I'd get to 3x4 + 2x1 just the same.”

With this method you don’t have to fiddle in this manner. You don’t have to try 3 times this, and 1 times that, with different factor pairs, in different permutations. All you have to do is choose the pair which add to give the middle term then plug them in to your brackets.

The only extra thing you then have to do is cancel where you can; in this case cancel (3x+12) to make it (x+4)

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Factorising a quadratic expression where the a term is greater than 1

@Bob
“Can you show me an example where it makes a real difference?”
I believe it makes a real difference in your example, but I’ll give another in case I’ve not been clear.
4x^2 -19x + 12
1.    Multiply the a term by the c term, i.e, 4x12=48
2.    Choose a factor pair from 48 that will add to give the b term, the middle term, i.e, -19. That pair will be -3 and -16.
3.    Write out your bracket pair but put 4x in each pair instead of 4x in one, and x in the other.
4.     Insert your factor pair, to give:
(4x-3)(4x-16)
5.    Cancel where you can, i.e, (4x-16) cancels to (x-4)
6.    Which gives: (4x-3)(x-4), which is your answer.
7.    Nb. The only extra step is the cancelling which takes less time than fiddling with the factor pair.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,196

Re: Factorising a quadratic expression where the a term is greater than 1

If a quadratic is in the form

Then it's 'easy' to factorise as (x-a)(x-b).

Your method takes advantage of this in the following manner.

Without the 4 this new quadratic has the above  form so we just have to find factors of 3 that make -19/4 **

-16/4 and -3/4 works so we get

But the ** step is made tricky as we're working with fractions, so not dividing by 4 too early leaves us with whole numbers and so it becomes easier.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Factorising a quadratic expression where the a term is greater than 1

Thanks, Bob.

But that level of algebra is too difficult for me.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,196

Re: Factorising a quadratic expression where the a term is greater than 1

hi paulb203

I'm a maths teacher so I ought to be able to do something about that.  I find a diagram often helps.  Let's start with the simplest case, a quadratic where the x squared term is just x^2.

If you create a random quadratic you'll probably not get a simple factorisation with whole numbers but if there is a simple factorisation then here's how we work out what it is.

Look for a pair of numbers that multiply to give Q, and add to give P.

ie. ab = Q and a + b = P

Then the factorisation is (x+a)(a+b). This picture shows why it works.

The two x terms add up like this: ax + bx = (a+b)x.

Does that make sense? If so I'll make the more complex (but not much!) diagram for when the first term is ax^2.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob