# factoring trinomial squares

• Jul 12th 2008, 03:54 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, 04:08 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

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

Let \$\displaystyle y^2 = u\$

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

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

Factor out a \$\displaystyle 5\$ from the first bracket.

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

\$\displaystyle =5(u + 1)^2\$

But \$\displaystyle u = y^2\$

Thus: \$\displaystyle =5(y^2 + 1)^2\$
• Jul 12th 2008, 04:10 AM
janvdl
Quote:

Originally Posted by mathtoe
heres another:

12p3 + 31p2 + 20p

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

Factor out a \$\displaystyle p\$.

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

\$\displaystyle = p(3p + 4)(4p + 5)\$
• Jul 13th 2008, 01:59 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, 02:08 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?

\$\displaystyle (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:

\$\displaystyle ( \ \ 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.