# factoring trinomial squares

• Jul 12th 2008, 04:54 AM
mathtoe
factoring trinomial squares
Hi, Im having difficulty factoring completely. Ive been running into this problem over again.

heres an example:

5y4 + 10y2 + 5

I ended up with:

5y(y3 + 1) 5(Y + 1)

5(y2 + 1)2

how do I know what Im supposed to completely factor? is there a rule Im fogetting? initially, I thought I was suppose to factor out all exponents but I found that not to be true.

heres another:

12p3 + 31p2 + 20p
• Jul 12th 2008, 05:08 AM
janvdl
Quote:

Originally Posted by mathtoe
Hi, Im having difficulty factoring completely. Ive been running into this problem over again.

heres an example:

5y4 + 10y2 + 5

I ended up with:

5y(y3 + 1) 5(Y + 1)

5(y2 + 1)2

how do I know what Im supposed to completely factor? is there a rule Im fogetting? initially, I thought I was suppose to factor out all exponents but I found that not to be true.

heres another:

12p3 + 31p2 + 20p

$5y^4 + 10y^2 + 5$

Let $y^2 = u$

$=5u^2 + 10u + 5$

$=(5u + 5)(u + 1)$

Factor out a $5$ from the first bracket.

$=5(u + 1)(u + 1)$

$=5(u + 1)^2$

But $u = y^2$

Thus: $=5(y^2 + 1)^2$
• Jul 12th 2008, 05:10 AM
janvdl
Quote:

Originally Posted by mathtoe
heres another:

12p3 + 31p2 + 20p

$12p^3 + 31p^2 + 20p$

Factor out a $p$.

$= p(12p^2 + 31p + 20)$

$= p(3p + 4)(4p + 5)$
• Jul 13th 2008, 02:59 AM
mathtoe
http://www.mathhelpforum.com/math-he...a5db08ea-1.gif

how did you arrive at this?

I was told to multiply the 1st & last terms and their sum equals the 2nd term. is that what you did? how else can you arrive with the 2 inner terms?
• Jul 13th 2008, 03:08 AM
janvdl
Quote:

Originally Posted by mathtoe
http://www.mathhelpforum.com/math-he...a5db08ea-1.gif

how did you arrive at this?

I was told to multiply the 1st & last terms and their sum equals the 2nd term. is that what you did? how else can you arrive with the 2 inner terms?

$(12p^2 + 31p + 20) = (3p + 4)(4p + 5)$

Basically you have to play around with the factors of 12 and 20.

For 12, you could have 12&1 ; 6&2 ; 3&4
For 20, you could have 20&1 ; 10&2 ; 4&5

Then you have to see which of those coefficients are capable of making 31 for the middle term.

So write:

$( \ \ p + \ \ )( \ \ p + \ \ )$ (We know both have to be + since the middle and last terms are positive.)

Now take those possible co-efficients, plug them in, and multiply out. If one set doesnt work, try the next.

Note that you can also swap the co-efficients around.
Instead of pairs 3&4 and 4&5 ; you could have 4&3 and 4&5.
Later on you will have the knack of instinctively knowing which it would probably be.

I can give you a hint, which is: The co-efficients will often be those close to each other. If we go back to the example of 12, it will most likely be 3&4 (or 4&3) that will work. But this is not always true.