# Thread: Need explanation for basic factoring...

1. ## Need explanation for basic factoring...

Hi guys,

I've just started a discrete mathematics course at university, and one of the things that is assumed knowledge is being able to factor expressions. The notes then go on to give you some exercises for you to test yourself with. There are accompanied answers, but they don't show how they worked them out. I've got some simple ones solved, but not the ones that are slightly more complicated

I haven't done anything like this for years and years, so I've forgotten it. I'll show you the example questions so hopefully someone can use them as a reference in explaining them to me.

Example:

Let $f(x)= 3x^{2} +2 ,
g(y) = 2y -1 and
h(z) = -z^{2} +3$
, express the following as polynomials:
a) f(y)
b) g(x)
c) h(n)
d) f(m+1)
e) g(n-2)
f) h(3k + 1)

The answers for the first three were $f(x)= 3y^{2} +2 ,
g(y) = 2x -1 and
h(z) = -n^{2} +3$
(really obvious actually, not really doing anything)

But the other three, I got halfway through but I couldn't find out how the final answer was arrived at. The final answers were given as these:
d) $f(m+1)= 3m^{2} + 6m +5$
e) $g(n-2) = 2n -5$
f) $h(3k + 1) = -9k^{2} - 6k + 2$

If I'm being at all vague (which I'm sure I am) please let me know and I'll try to reword the problem I'm having.

2. Replace $x$ with $m+1$ therefore,

$f(m+1)= 3(m+1)^2+2= 3(m^2+2m+1)+2= 3m^2+6m+3+2= 3m^2+6m+5$

do you follow?

3. Originally Posted by Hiram
Hi guys,

I've just started a discrete mathematics course at university, and one of the things that is assumed knowledge is being able to factor expressions. The notes then go on to give you some exercises for you to test yourself with. There are accompanied answers, but they don't show how they worked them out. I've got some simple ones solved, but not the ones that are slightly more complicated

I haven't done anything like this for years and years, so I've forgotten it. I'll show you the example questions so hopefully someone can use them as a reference in explaining them to me.

Example:

Let $f(x)= 3x^{2} +2 ,
g(y) = 2y -1 and
h(z) = -z^{2} +3$
, express the following as polynomials:
a) f(y)
b) g(x)
c) h(n)
d) f(m+1)
e) g(n-2)
f) h(3k + 1)

The answers for the first three were $f(x)= 3y^{2} +2 ,
g(y) = 2x -1 and
h(z) = -n^{2} +3$
(really obvious actually, not really doing anything)

But the other three, I got halfway through but I couldn't find out how the final answer was arrived at. The final answers were given as these:
d) $f(m+1)= 3m^{2} + 6m +5$
e) $g(n-2) = 2n -5$
f) $h(3k + 1) = -9k^{2} - 6k + 2$

If I'm being at all vague (which I'm sure I am) please let me know and I'll try to reword the problem I'm having.
$f(y)=3y^2+2$

$g(x)=2x-1$

$h(n)=-n^2+3$

$f(m+1)=3(m+1)^2+2=3(m+1)(m+1)+2=3[m(m+1)+1(m+1)]+2$

$g(n-2)=2(n-2)-1$

$h(3k+1)=-(3k+1)^2+3=-(3k+1)(3k+1)+3$ etc

4. Not really, where did the 2m in the second iteration come from?

5. $(m+1)^2=(m+1)(m+1)=m(m+1)+1(m+1)=m^2+m+m+1=m^2+2m+ 1$

Which is based on $5(5)=(3+2)(3+2)=3(3+2)+2(3+2)=3^2+3(2)+2(3)+2(2)$ and so on...

6. I guess what I'm getting at is, I don't know how to factorise. Thanks anyway guys, I'll find someplace else.

7. Originally Posted by Hiram
I guess what I'm getting at is, I don't know how to factorise. Thanks anyway guys, I'll find someplace else.
These questions are not about factorisation, but about substituting one expresion (infact a linear expresion) for the variable in another (a polynomial in these cases).

8. OP: be careful with the terminology. It's not factoring, it's just multiplying. Factoring is going backwards, in a sense, where you are given a product and you need to find two (or more numbers) multiplied together to get to that product. So doing something like this:
$(x + 3)(x + 2) = x^2 + 6x + 5$
...which is what you say you don't know, is multiplying, but doing something like this:
$x^2 - 8x + 15 = (x - 3)(x - 5)$
... is factoring. See the difference?