# Thread: help with specific trinomials.

1. ## help with specific trinomials.

right, so i know how to do trinomials in a normal case with only one pronumeral.

a^2 + 2a + 1
(a+1)(a+1)

and i can do non-monic trinomials no problem too, problem for me is when i discover a trinomial which has 2 pronumerals.

a^2 + 2ab + b^2
now i have no idea how i'm meant to work it out.
i know that this simplifies to (a+b)(a+b) because i can do the expansion for it.
but i have no idea how to simplify it and if it were slightly more complex
eg. k^2 + 8kp + p^2 i would not know what to do in the exam.

help please! could someone please show me how to work out these expressions a^2 + 2ab + b^2

2. Originally Posted by Zellus
right, so i know how to do trinomials in a normal case with only one pronumeral.

a^2 + 2a + 1
(a+1)(a+1)

and i can do non-monic trinomials no problem too, problem for me is when i discover a trinomial which has 2 pronumerals.

a^2 + 2ab + b^2
now i have no idea how i'm meant to work it out.
i know that this simplifies to (a+b)(a+b) because i can do the expansion for it.
but i have no idea how to simplify it and if it were slightly more complex
eg. k^2 + 8kp + p^2 i would not know what to do in the exam.

help please! could someone please show me how to work out these expressions a^2 + 2ab + b^2

Complete the square:

$\displaystyle k^2 + 8p + p^2 = (k + 4p)^2 - 15p^2 = (k + 4p)^2 - (\sqrt{15} p)^2$.

Now use the difference of two squares formula to get $\displaystyle k^2 + 8p + p^2 = (k - [\sqrt{15} - 4] p) (k + [ \sqrt{15} + 4] p)$.

3. Originally Posted by mr fantastic
Complete the square:

$\displaystyle k^2 + 8p + p^2 = (k + 4p)^2 - 15p^2 = (k + 4p)^2 - (\sqrt{15} p)^2$.
how did you know that k^2 + 8p + p^2 when has it's square completed = (k + 4p)^2 - 15p^2 without having to go through a slow, laborious process of trial and error?

4. Originally Posted by Zellus
how did you know that k^2 + 8p + p^2 when has it's square completed = (k + 4p)^2 - 15p^2 without having to go through a slow, laborious process of trial and error?
The key is to know that $\displaystyle {\color{red}A^2 + 2AB} + B^2 = (A + B)^2$. So $\displaystyle {\color{red}k^2 + 8p} + 16p^2 = (k + 4p)^2$.

Hence $\displaystyle k^2 + 8p + p^2 = k^2 + 8p + 16p^2 - 15p^2 = (k + 4p)^2 - 15p^2$.