# Thread: Need help factoring these equations!

1. ## Need help factoring these equations!

I have a couple of questions actually, I have a test tommorow on factoring from standard form PLEASE HELP ! It would be much appreciated.
so here are the problems I'm having difficulties with.
a)6h^2 -17+ 12 b)24a^2b^3 + 9a^2 + b^2 + 18a^3b^2

c) 4k^2 + 36k + 81 d) 4p^2q - 8p^2 - q +2

Question 2 : 9x^2 + Kx +144 - This is a perfect square trinomial what is the value of k? explain.

Factor :
1) a^2 - 2a + 1 - b^2 + 2bc - c^2
THANK YOU SO MUCH TO ANYONE THAT ANSWERS!!!

2. Originally Posted by farfromover

Question 2 : 9x^2 + Kx +144 - This is a perfect square trinomial what is the value of k? explain.

For this to be a perfect square then $9x^2 + Kx +144 = (3x+12)(3x+12)$

Expand the RHS and you will find $K$

You really need to show some workings so we know where to help. Giving you answers in the 11th hour before a test is not what you need. You need to learn how to apply the techniques of factorisation.

3. Hello, farfromover!

1) Factor: . $a^2 - 2a + 1 - b^2 + 2bc - c^2$

We have two "groups": . $(a^2-2a+1) - (b^2-2bc+c^2)$

Factor the groups: . $(a-1)^2 - (b-c)^2$ . . . . difference of squares

. . . . $=\;\bigg[(a-1)-(b-c)\bigg]\,\bigg[(a-1) + (b-c)\bigg]$

. . . . $=\;(a-1-b+c)\,(a-1+b-c)$

4. thank you!

5. so is K equal to 0?

6. Originally Posted by farfromover
so is K equal to 0?
Show me the expansion.

7. Originally Posted by pickslides
For this to be a perfect square then $9x^2 + Kx +144 = (3x+12)(3x+12)$

Expand the RHS and you will find $K$

You really need to show some workings so we know where to help. Giving you answers in the 11th hour before a test is not what you need. You need to learn how to apply the techniques of factorisation.
Expansions: 9x^2 + 36x + 36x + 144
= 9^x + 72 + 144
Oh so its equal to 72.
can the 72 be negative also?
Thank you

8. Originally Posted by farfromover
can the 72 be negative also?
Yes, this will give $(3x-12)(3x-12)$

9. thank you so much, and what is this one its tricky :
9 - (5n-1) ^2

I tried letting x = (5n-1) but when i check it its incorrect

10. Differnece of two squares says $a^2-b^2 = (a-b)(a+b)$

$9 - (5n-1) ^2 = 3^2 - (5n-1) ^2 = (3-(5n-1))(3+(5n-1))= \dots$

simplify to finish..