# I need help again also would like to see if someone can double check for me

• Sep 11th 2005, 05:46 PM
UKMUG
I need help again also would like to see if someone can double check for me
These are the ones I could use help with:
Reduce to lowest term:
x^2-y^2
------------- (fraction)
x^2+2xy+y^2

X^2-6x+9 3
---------- X ----- (fraction)
3x x-3

Find the LCD by factoring

4 3
------ , --------
x^2-16 x^2+8x+16

If the length of a rectangle is 5/2x feet and the width is 2/3x feet, express the perimeter of the rectangle in the simplest form? :

Check:

Factor out the GCF:
6x^4y^2-18x^2y^2+9x^2y^4

Factor:
16x^2-49y^2

Factor:
x^2+6x-16

Factor:
X^3-8

7 2
-------- + --------
2x-6 x-3
---------
2x^2-12x+18

• Sep 11th 2005, 11:55 PM
hoeltgman
1.$\displaystyle$ \dfrac{x-y}{x+y} 

2.$\displaystyle$ \dfrac{x-3}{x} 

3.$\displaystyle$ \text{LCD:} \; (x+4)^2\cdot(x-4) 

4.$\displaystyle$ \dfrac{19}{3} \cdot x 

5.$\displaystyle$ 3x^2\cdoty^2\cdot(2x^2 + 3(y^2 - 2)) 

7. $\displaystyle$ (x - 2)\cdot(x + 8) 

9. $\displaystyle$ \dfrac{11}{2(x-3)} 
• Sep 12th 2005, 11:31 AM
hemza
In details
Before beggining, lets first see the 2 main methods of factorisation:
1) subtraction of two squares : if you have for example 9 x^2 - 25 y^2, 9x^2 and 25 y^2 are the squares of 3x and 5y and because there is a minus sign between 9x^2 and 25 y^2, you can use a^2 - b^2 = (a-b)(a+b) where a=3x and b=5y so 9 x^2 - 25 y^2 = (3x-5y)(3x+5y).
To verify that it is right, we multiply (3x-5y)(3x+5y) so (3x-5y)(3x+5y) = 9x^2 +15xy - 15xy - 15y^2 = 9x^2 - 15y^2. So it is good.

2) trinom : if you have for example 3 x^2 -2x - 5 (polynom of the form ax^2+bx+c), what we do is find two numbers m and n such that m*n=3*(-5)=-15 (m*n = a*c) and m+n = -2 (m+n = b). Make sure you've got the right signs !!! Here, we find that m =3 and n=-5 (or m=-5 and n=3 it is not important). So since we know b=-2=m+n we replace b by m+n and get :
3x^2 + (3-5)x - 5 = 3x^2 + 3 x -5x - 5. We factorize each the two first terms together by finding the GCD and the two last together by the same method.
So, 3x^2 + 3 x -5x - 5 = 3x(x+1)-5(x+1) and we factorize the common factor (x+1) out. So we end up with (3x-5)(x+1). Done!
So 3x^2 -2x - 5 = (3x-5)(x+1).
To verify that it is right, we multiply (3x-5)(x+1) so (3x-5)(x+1) = 3x^2 + 3x - 5x - 5 = 3x^2 -2x-5.
Ok ? There are many shortcuts for this (2nd) method when you have special numbers but you should practice it this way (the long way) because this method works every time you can factorize a trinom.

1. (x^2 - y^2) / (x^2 + 2xy + y^2) = [(x-y)(x+y)]/[(x+y)(x+y)] = (x-y)/(x+y). I used the 1st method for (x^2 - y^2) and the second for the denominator.

2.(x^2 - 6x + 9) /(3x) * 3/(x-3) multiplication of fractions goes like : (numerator1*numerator2) /(denominator1 * denominator2) so : [3(x^2 - 6x + 9)]/[3x * (x-3)] = [(x^2 - 6x + 9)]/[x * (x-3)] = [(x-3)(x-3)]/[x * (x-3)] = (x-3)/x
Here I used the 2nd method for factorizing (x^2 - 6x + 9).

3. 4/(x^2 - 16) = 4/[(x-4)(x+4)] (by 1st method) and 3/(x^2+8x+16) = 3/[(x+4)(x+4)] (by 2nd method) so the LCD is (x+4)^2 (x-4)

4. p = 2 (L+W) = 2(5/2 x + 2/3 x). You add the fractions 5/2 with 2/3 so p = = 2(19/6 x) = 19/3 x.

5. 6x^4 y^2-18x^2 y^2+9x^2 y^4 So we have everywhere a 3, an x^2 and a y^2 so 3x^2 y^2 (2x^2 - 6 + 3y^2).

6. 16x^2 - 49 y^2 (1st method) so it is (4x)^2 - (7y)^2 = (4x-7y)(4x+7y)

7. x^2 + 6x -16 (2nd method) so two numbers m and n such that m*n=1*(-16) = -16 and m+n=6 so m=8 and n=-2 (or m=-2 and n=8). We replace 6x by (8-2)x and get x^2 + 8x - 2x -16 = x(x+8)-2(x+8) = (x-2)(x+8).

8. x^3 - 8. Here, we do not have the 1st method because x^3 and 8 are not squares. In this kind of problem you either know the shortcut or use the following technique. So we use a technique which goes: since we have -8 at the end we guess that x^3-8 should be divisible by factors of -8. So you try small factors h at the beggining until the division (x^3 - 8)/(x-h) is complete (no rest). Let's try -2. (x^3 - 8) /(x-2) = (x^2 + 2x+4) with no rest so it is good. So x^3-8 = (x-2)(x^2 + 2x+4) and we try to factorize (x^2 + 2x+4) with the 2nd method, we see that it is impossible to find m and n such that m*n=4 and m+n=2 so we cannot factorize it. So the answer is x^3-8 = (x-2)(x^2 + 2x+4)

9. 7/(2x-6)+2/(x-3). So we factorize the first denominator and get 7/[2(x-3)]+2/(x-3). So the LCD is 2(x-3). So 7/[2(x-3)]+4/[2(x-3)]. (we multiply the 2nd fraction with 2 at the nominator and the denominator). Now, we add. 11/[2(x-3)]. Ok.

Done!