Factorising

Write

4x^4-13x^2+9

as the product of four linear factors.

I understand that I most likely need to make this into a quadratic expression. In this particular instance, though, I cannot find a way to do this. I cannot find a common factor between

4x^4

-13x^2

and

9

. I have written expressions as the product of three linear factors before, but not four.

(edited 6 years ago)

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Reply 1

B_9710

Notice we have

4(x^2)^2-13(x^2)+9

so if we let

y=x^2

then we just have a quadratic in y.
Just factorise this as normal and then you should be able to factorise further.

Reply 2

Dysf(x)al

Original post by Illidan2

Write

4x^4-13x^2+9

4x^4

-13x^2

and

9

. I have written expressions as the product of three linear factors before, but not four.

Because there are no

x^3

x

terms, you can let

y=x^2

, then you just have a regular quadratic. Factorise it into two factors, then substitute the

x^2

back in place of y. Then you should be able to factorise each of those two bits into two factors each by using the difference of two squares method (

a^2 - b^2 = (a+b)(a-b)

Reply 3

Illidan2

Thanks. I am quite new to this, so I don't know if i'm on the right track. Unsure of whether I have followed your method correctly so far, I now have two brackets,

4(y^2-1)

and

9(1-1)

. Does that make sense so far? Or have I gone wrong already? If that's correct so far, how do I go about substituting the

x^2

back in? Would it just be

4((x^2)^2 -1)

(edited 6 years ago)

Reply 4

Quirky Object

Original post by Illidan2

Thanks. I am quite new to this, so I don't know if i'm on the right track. Unsure of whether I have followed your method correctly so far, I now have two brackets,

4(y^2-1)

and

9(1-1)

. Does that make sense so far? Or have I gone wrong already? If that's correct so far, how do I go about substituting the

x^2

back in? Would it just be

4((x^2)^2 -1)

I'm having trouble seeing how you got that. What do you get when you factorise 4y^2 - 13y + 9, first of all?

Reply 5

Illidan2

ac=36.

-9-4=-13=b

4y^2 -9 -4 +9 = (4y-3)^2 (-2 +3 )^2

(4y-3) (4y-3) (-2 +3) (-2+3)

(edited 6 years ago)

Reply 6

Quirky Object

Original post by Illidan2

ac=36.

-9-4=-13=b

4y^2 -9 -4 +9 = (4y-3)^2 (-2 +3 )^2

(4y-3) (4y-3) (-2 +3) (-2+3)

I can't say I've ever seen that method of factorising a quadratic before.

-2 + 3 = 1

, so you've just ended up with

(4y + 3)^2

and

(4y + 3)^2 \neq 4y^2 - 13y + 9

. Also there should still be a

y

in your second line. How would you have factorised a quadratic at GCSE level?

Reply 7

Notnek

Original post by Illidan2

ac=36.

-9-4=-13=b

4y^2 -9 -4 +9 = (4y-3)^2 (-2 +3 )^2

(4y-3) (4y-3) (-2 +3) (-2+3)

This looks at an attempt at the AC method but then where has

(-2+3)^2

come from?

Or like @Sonechka , I may just not be familiar with this method. But you haven't arrived at the correct answer anyway.

Reply 8

Illidan2

Yes, that was a mistake. 9 and 4 should have been

9y-4y

I'm unsure as to how I would have done it at GCSE level, it has been over 5 years since I sat my GCSE exams. Until around 2 days ago, I was not sure I had factorised a quadratic since.

(edited 6 years ago)

Reply 9

Notnek

Original post by Illidan2

Yes, that was a mistake. 9 and 4 should have been

9y-4y

I'm unsure as to how I would have done it at GCSE level, it has been over 5 years since I sat my GCSE exams. Until around 2 days ago, I am not sure I had factorised a quadratic since.

It seems likely that you are trying to use the AC method. Try watching this video then have another go.

You'll need to be able to factorise quadratics before you can tackle the question in your OP.

Reply 10

Illidan2

Yes, that's exactly right. I'm trying to use the AC method, as it is the only method I know for factorising quadratics, which is what I read in a textbook 2 days ago. I didn't know other methods existed. I'll watch the video, then give it another try, like you said.

Reply 11

Notnek

Original post by Illidan2

I didn't know other methods existed. I'll watch the video, then give it another try, like you said.

There are 3 methods that I can think of. Two of them (including the AC method) are taught maybe 50-50 in schools. Most students learn this topic without realising that half the country does it differently :smile:

Reply 12

Illidan2

4y(y+1) -9(y+1)

(y+1) (4y-9)

I realise now where I went wrong. Now that I managed to factorise that quadratic, I'm not sure how to put the x^2 back in in place of y and complete it.

Edit: I apologise for the errors in my recent posts, I've had to switch to my phone.

(edited 6 years ago)

Reply 13

Quirky Object

Original post by Illidan2

4y(y+1) -9(y+1)

(y+1) (4y-9)

You've made a sign error; remember c = +9, so you need -1 and -9 in the brackets or you'll end up with -9 or +13 instead of +9 and -13. It's always a good idea to check your factorising by expanding out the brackets again.

You can now write the factorised expression as

(x^2-1)(4x^2 - 9)

. What do you notice about

1, 4, 9

and

x^2

Reply 14

Illidan2

Sorry for the delayed reply, I'm at work you see, only just got on break.

As far as I understand it, 1, 4, 9 and

x^2

are each, individually, linear factors. Seeing as there are four or them, I assume that is the problem written out as a product of four linear factors? :smile:

So effectively, as other people in this thread have said, this is just a case of ordinary quadratic factorisation with some additional substitution. I see! It all becomes clear now :smile:

(edited 6 years ago)

Reply 15

Illidan2

Suppose I have a similar problem, with

a^4

but just

b

. What would I do then? Would substitution still work? It seems like it wouldn't be possible to substitute as you can't take x^2 off both sides to convert it to a quadratic.

Reply 16

Quirky Object

Original post by Illidan2

Sorry for the delayed reply, I'm at work you see, only just got on break.

As far as I understand it, 1, 4, 9 and

x^2

are each, individually, linear factors. Seeing as there are four or them, I assume that is the problem written out as a product of four linear factors? :smile:

So effectively, as other people in this thread have said, this is just a case of ordinary quadratic factorisation with some additional substitution. I see! It all becomes clear now :smile:

Yes, that is what it is, but you're not quite there. By "four linear factors", the question means "four sets of brackets", essentially.

The next step relies on you noticing a numerical property of 1, 4, 9 and