Factoring rules, I have the anwser but im looking for how to do it please help?

Question

Given: 9y^2-(2x-y)^2

Book Anwser: -4(X+Y)(X-2Y)

GIVEn: 25^2-(x+2y+4z)^2

Anwser: (X+a+b)(x-a-b)

Please explain in steps of how to factor the Given problems to the arrival of the anwsers.
Explain in steps PLEASE! THANK YOU

Answer 1

Hi,

These problems are dealing with the difference of perfect squares. The general formula for this type of problems is:

K^2 - L^2 = (K - L)(K + L)

Given: 9y^2-(2x-y)^2

In this problem 9y^2 is in the place of K^2. You would square 3y to get this term. (3y * 3y = 9y^2)
Likewise (2x-y)^2 is in place of L^2, so L is really (2x - y) in this problem.

So 9y^2-(2x-y)^2 factors into:

(3y - (2x - y))(3y + (2x - y))
(..K.-.....L.....)(.K..+....L.....) See the pattern?

Now simplify each parentheses by eliminating inside parentheses and combining like terms:

(3y - (2x - y))(3y + (2x - y))
(3y - 2x + y)(3y + 2x - y)
(4y - 2x)(2y + 2x)

You can factor a 2 out of each parentheses:
(4y - 2x)(2y + 2x) =
2(2y - x)2(y + x)

They multiply together to give your final answer on the first problem of:

4(2y - x)((y + x)

If you want the x terms first, factor a "-1" out of the first parentheses. Reverse the order in the parentheses with a +.

4(-1)(x - 2y)((x + y) =
-4(x - 2y)((x + y) This matches your book answer for the first problem.




Next problem

Use the pattern: K^2 - L^2 = (K - L)(K + L)

Given: 25^2-(x+2y+4z)^2


In this problem 25^2 is in the place of K^2. So K = 25.
Likewise (x + 2y + 4z)^2 is in place of L^2, so L is really (x + 2y + 4z) in this problem.

So 25^2-(x+2y+4z)^2 factors into:

(5 - (x + 2y + 4z))(5 + (x + 2y + 4z))
(..K.-.....L.....)(.K..+....L.....) See the pattern?

Now simplify each parentheses by eliminating inside parentheses:

(5 - x - 2y - 4z)(5 + x + 2y + 4z)

This is your second problem's answer. This doesn't match the book's answer. That answer with "a" and "b" in it doesn't make any sense when this problem doesn't have those letters. I think you looked at the wrong answer!


I hope this helps you!! :-)

Answer 2

9y^2-(2x-y)^2
first expand the second expression:
9y^2 -4x^2 +4xy-y^2 =
8y^2 +4xy -4x^2=
-4(x^2 -xy -2y^2)
-4(x+y)(x-2y)
this is done by inspection or trial and error you need 2 factors such that the product of the first two terms is the x^2, the sum of the products of the 1st and second and the second and the first is xy and the product of the second times the second is 2y^2. in the present example
x*x=x^2
x*-2y +x*y= -xy
y*y = y^2

your second problem is unclear the given expression is in terms of x,y and z while the answer is in terms of x,a and b

Answer 3

The first thing you want to do is group like terms together. So for 9y^2 -(2x-y)^2 multiply it out and then group like terms:

9y^2 - ((2x-y)(2x-y))
9y^2 - (4x^2-2xy-2xy+y^2)
9y^2 - 4x^2 + 4xy - y^2
8y^2 + 4xy - 4x^2

Note that you can factor out -4 from all terms
-4(-2y^2 - xy + x^2)

Now consider the terms in the bracket. To get -2y^2 you need -2y times y. To get x^2 you need x times x. and to get -xy you need -2xy+xy. Therefore
(-2y+x)(y+x) so you get

-4(x-2y)(x+y)

Answer 4

Given: 9y^2-(2x-y)^2
=(3y)^2-(2x-y)^2
using a^2-b^2 =(a+b)(a-b)
we get for the given expression: (3y+2x-y)(3y-2x+y)=(2x+2y(4y-2x)=4(y+x)(2y-x).OK.
25^2-(x+2y+4z)^2(25+x+2y+4z)(25-x-2y-4z).

Answer 5

Question 1
[ 3y - (2x - y) ][ 3y + (2x - y) ]
(4y - 2x).(2y + 2x)
2.(2y - x).2(x + y)
- 4.(x + y).(x - 2y)

Question 2
[ 5 - (x + 2y + 4z) ].[ 5 + ( x + 2y + 4z) ]

Answer 6

=>9y^2-(4x^2-4xy+y^2)=9y^2-4x^2+4xy-y^2
=8y^2-4x^2+4xy
take -4 as common factor
v get -4(x^2-xy-2y^2)= -4(x^2-2xy+xy-2y^2)
in the paranthesis take x and y as common factor
-4(x(x-2y)+y(x-2y))= -4(X+Y)(X-2Y)

similarly solve de 2nd 1...

jj

Answer 7

a^2 - b^2 = (a-b)(a+b)
In your first problem a=3y and b=(2x-y)

(3y-2x+y)(3y+2x-y) = (4y-2x)(2y+2x)

Take 2 out from BOTH
2(2y-x)2(y+x) = 4(2y-x)(y+x) = -4(x-2y)(y+x)

Same logic for second problem

Answer 8

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