How do I factorise polynomials using a new method?

[Hawksteinman]

As part of my degree in physics with Astrophysics, I have to do some maths modules. In the maths lectures, the lecturer just goes though a giant 212 page work booklet explaining everything as she goes along. Me and a friend only do the work booklet in the lectures and we're already on page 33 while everyone else is on page 14.

I am not sure if I have done these questions correctly (page 21), because I've never used this method of factorising polynomials before (there was an example of how to do it in the workbook)

 

[Hawksteinman]

PS that's a z I'm using as another variable

 

[mathman]

The image is very hard to read. The bottom 3 lines look OK, but I can't decipher the rest.

 

[DS2C]

Very hard to read!
Does the first one read ##3x^2+11x-4##?
If so your factorization is incorrect.
3 is not a GCF of the trinomial and you would have to look at the factors of ##3x^2## and ##-4##.
You can check the solution by FOILing it back into the original expression.

 

[Hawksteinman]

Very hard to read!
Does the first one read ##3x^2+11x-4##?
If so your factorization is incorrect.
3 is not a GCF of the trinomial and you would have to look at the factors of ##3x^2## and ##-4##.
You can check the solution by FOILing it back into the original expression.

It is, but I was using the method we were told to use where you multiply the whole expression by the coefficient :/

 

[DS2C]

You are probably in a higher math than I, but we are doing factorization right now.
As a helpful guide, here's a pic of my notes that relates to the first problem. Try and apply them to your first problem. Hopefully they're legible enough.

 

[bluenoise]

I'd say, the method you are trying to apply is a kind of substitution. You first multiply the polynomial with the coefficient of the ##x^2## term, so you get:
##(5x)^2+17\cdot(5x)+30##
Now substitute a new variable, say u, for 5x
##u^2+17u+30##
Factorise by educated guessing (the product of the two numbers must be 30, the sum must be 17) :
##(u+2)(u+15)##
Substitute the 5x back:
##(5x+2)(5x+15)##
Factorise the second parenthesis:
##(5x+2)(x+3)\cdot5##
Discard the factor 5 and you are left with the solution:
##(5x+2)(x+3)##

 

[Mark44]

Two things:

1. The image is extremely difficult to read. If you are going to post an image (which is generally frowned on here because so many of them are just as unreadable as yours), at least make sure that the work is properly lit before taking a photo.
2. What did you multiply by 3 right away? That makes no sense. The factorization problem you posted can be done in two lines:
   ##3x^2 + 11x - 4##
   ##= (3x - 1)(x + 4)## Done.

Everything you have in the photo can be done using either BBCode or (better) LaTeX. We have tutorials on each of these. On the menu bar across the top, one of the choices is INFO. Under this item is a drop -down menu with a choice of Help/How-to. The BBCode and LaTeX tutorials are listed under this menu choice.

Finally, I moved your thread to General Math, as it had nothing to do with Linear Algebra or Abstract Algebra.

 

[mfb]

It is, but I was using the method we were told to use where you multiply the whole expression by the coefficient :/

If you have something like ##3(x^2+4x-5)##, sure, but you don't have that.
What you could do here would be ##3x^2 +11 x -4=3\left(x^2+ \frac{11}{3} x - \frac{4}{3}\right)##, but that doesn't help.

You can't just multiply the expression by 3, that changes its value (instead of just how it is presented).

 

[Hawksteinman]

If you have something like ##3(x^2+4x-5)##, sure, but you don't have that.
What you could do here would be ##3x^2 +11 x -4=3\left(x^2+ \frac{11}{3} x - \frac{4}{3}\right)##, but that doesn't help.

You can't just multiply the expression by 3, that changes its value (instead of just how it is presented).

Yes and you divide it by three again at the end :)

 

[Mark44]

Yes and you divide it by three again at the end :)

Seems like wasted effort if you multiply by 3 in the beginning, and then divide by 3 at the end. This might be useful to do if there was a point to it, but here there isn't.

 

[bluenoise]

Seems like wasted effort if you multiply by 3 in the beginning, and then divide by 3 at the end. This might be useful to do if there was a point to it, but here there isn't.

Of course there is. It is done in order to substitute 3x with another variable, which reduces the complexity of the original term. A substititution/back-substitution method as used for similar purposes elsewhere... maybe not really needed for this kind of problem, I agree, but usefull still.

 

[mfb]

I don't see the point, but then you still need to keep the factor 1/3 in the equation. Otherwise it is not an equation any more. You can't just say "5=3*5" because you want a factor 3, and what was done here is equivalent.

 

[bluenoise]

That's why one has to divide by 3 in the end. See my post #7.

 

[sirios]

The image is very hard to read. The bottom 3 lines look OK, but I can't decipher the rest.

I agree

 

[mfb]

That's why one has to divide by 3 in the end. See my post #7.

5=3*5 doesn't get right just because you write 3*5=5 somewhere later (which is wrong as well).
You can write 5=1/3 * 3 * 5 = 1/3 * (whatever) = ... = 5, that is correct.