Factoring-can't find the right factors

[Simonio]

I'm trying to factorize this:

\(16x^2 -144x+99\)

The only factors I can find of ac are: 132,12 which add up to -144 but only if both are negative and we need opposing signs to get the -144. I've used an online factor calculator and can't seem to find anything there! I'm obviously missing something obvious , so any help welcome!

Whoops, should be 16x squared of course-don't know why it isn't coming out right!

 

[MarkFL]

We want two factors of $16\cdot99=1584$ whose sum is $-144$. Well, we may observe that:

$16\cdot99=(4\cdot3)(4\cdot33)=12\cdot132=1584$

And so we know:

\(\displaystyle (-12)(-132)=1584\)

So, you have found the correct factors...they will add up to -144 and their product will be positive. Can you put this together to get the factored form?

 

[Simonio]

Thanks(Clapping)-so now I can say:

\(16x^2-132x-12x + 99\) Then factorise in two groups:

\(4x(4x-33)(-4x+33)\) then

\((4x-3)(-4x-33)\)

So: \(x = -\frac{-3}{4}\) or \(x = \frac{33}{4}\)

Have I used the best method here?

 

[Deveno]

Thanks(Clapping)-so now I can say:

\(16x^2-132x-12x + 99\) Then factorise in two groups:

\(4x(4x-33)(-4x+33)\) then

\((4x-3)(-4x-33)\)

So: \(x = -\frac{-3}{4}\) or \(x = \frac{33}{4}\)

Have I used the best method here?

This isn't right. Let's start over:

$16x^2 - 144x + 99 = 16x^2 - 132x - 12x + 99$

$= (4x)(4x) - (4x)(33) - (3)(4x) + 3(33) = 4x(4x - 33) - 3(4x - 33)$

You must apply the distributive law correctly. Can you continue from here?

 

[MarkFL]

You may find our tutorial on factoring quadratics useful:

http://mathhelpboards.com/math-notes-49/factoring-quadratics-3396.html

 

[Simonio]

Thanks (i'm trying to teach myself after many years of not doing maths so this help is invaluable):

So \(4x(4x-33)-3(4x-33)

Then: \((4x-3)(4x-33)\)

So: \(x=\frac{3}{4}\) or \(x=\frac{33}{4}\)

Is this now ok? I can see I had been sloppy before-I'm finding it's so easy to make small mistakes. I think I'm going to need a lot of practise!

 

[MarkFL]

Thanks (i'm trying to teach myself after many years of not doing maths so this help is invaluable):

So \(4x(4x-33)-3(4x-33)

Then: \((4x-3)(4x-33)\)

So: \(x=\frac{3}{4}\) or \(x=\frac{33}{4}\)

Is this now ok? I can see I had been sloppy before-I'm finding it's so easy to make small mistakes. I think I'm going to need a lot of practise!

Yes, that's correct. Another way you could look at it is to let $u=4x$ and you have:

\(\displaystyle u^2-36u+99\)

Now we have the easier task of finding two factors of 99 whose sum is -36, which are -3 and -33:

\(\displaystyle u^2-36u+99=(u-3)(u-33)\)

Then back-substitute for $u$ and write:

\(\displaystyle (4x-3)(4x-33)\)

 

[Deveno]

Thanks (i'm trying to teach myself after many years of not doing maths so this help is invaluable):

So \(4x(4x-33)-3(4x-33)

Then: \((4x-3)(4x-33)\)

So: \(x=\frac{3}{4}\) or \(x=\frac{33}{4}\)

Is this now ok? I can see I had been sloppy before-I'm finding it's so easy to make small mistakes. I think I'm going to need a lot of practise!

Practice is good: it's how we make ourselves comfortable with new (or long unused) ideas.