How do i factorise the below?

How do i factorise the below? (a+2)(2a)+7a+14 and 64x^2+80xy+25y^2

First one:


second:

 
We'll use grouping - there are many ways...but grouping works well for this problem 
 
 
 
 
or 
 
which leads to 
v=1/3 
or v=1

Factorise xy^2-3xy-10x

xy^2-3xy-10x=
x(y^2-3y-10)=
x(y-5)(y+2)

Factor: 15y^2+8y+1

since 3*5=15 and 3+5=8...


15y^2+8y+1=
15y^2+3y+5y+1=
(15y^2+3y)+(5y+1)=
3y(5y+1)+1(5y+1)=
(3y+1)(5y+1)

factor: 5x(2x-5)^2 - (2x-5)^3

5x(2x-5)^2 - (2x-5)^3= 
(2x-5)^2(5x-(2x-5))= 
(2x-5)^2(3x+5)

factor: x3 - 12x2 + 48x - 64

x3 - 12x2 + 48x - 64 =
x3-64 -12x(x-4)=
(x-4)(x2+4x+16)-12x(x-4)=
(x-4)(x2-8x+16)=
(x-4)3 

:)

±√324 - how do you solve this

A method of extracting roots is to use the factorization of a number (by successively dividing out primes until the number is manageable) 
factoring 324... 
324=2*162=
2^2*81=2^2*9^2 

so that 324 = (2*9)^2 = 18^2 

therefore the square root of 324 is 18 
±√324 = ±18

:)

solve: 8x^2-3=2x

8x^2-3=2x 
8x^2-2x-3=0 
8x^2+4x-6x-3=0 
(8x^2+4x)-(6x+3)=0 
4x(2x+1)-3(2x+1)=0 
(4x-3)(2x+1)=0 

x=3/4 or x=-1/2

factorize: 25x^3-10xy^2+xy^2

 
 
:)

Factor: (3x+2)^2+8(3x+2)+12

12 = (6)(2) and 6+2 = 8 so we can write: 

 
then... 

 
so that means 

 

:)

Find the smallest number by which 37044 should be multiplied so that the product is a perfect cube

 
so multiply by 2 

Find the smallest number by which 6048 should be multiplied so that the product is a perfect cube

 
so multiply by 98 

Find the smallest number by which 2304 should be multiplied so that the product is a perfect cube

 
so multiply by 3 

How do you factor: 2x^2+ xy- y^2+ 3x+ 1

2x^2+ xy- y^2+ 3x+1= 

(2x-y)(x+y)+3x+1= 

(2x-y)(x+y)+2x+x+1= 

(2x-y)(x+y)+(2x-y)+(x+y+1)= 
(2x-y)(x+y+1)+(x+y+1)= 

(2x-y+1)(x+y+1) 


:)

Factor: 4x^2 - 16x - 9

4x^2-16x-9=0 


4x^2 - 16x = 9 
4(x^2-4x)=9 

4(x^2-4x+4)=9+4(4) 
4(x-2)^2=25 
(x-2)^2=25/4 

x-2=5/2
2x-4=5
2x-9=0 
or 

x-2=-5/2
2x-4=-5
2x+1=1 


(2x-9)(2x+1)=0

Factorise: (x - 1)^2 - 4y^2

Solve: 16x^(3)-100x=0

x=0
4x=10, x=5/2
4x=-10, x=-5/2


:)

Factor the expression: a^4b^2+ab^5

 


 

 



:)

one of my posts on Algebra.com

Solve: 231=7(x+8)x

231=7(x+8)x
the 231 seems large, so we look for a way to simplify - and it turns out to be a multiple of 7:



using cancellation and rearranging terms we get the following:




Notice that 11+(-3)=8 and (11)(-3)=-33 so we'll use them in our grouping:








which has solutions

x=3 and x=-11

:)

Factor