Factorization

Jan 2009
37
5
How do you factorize the following?

1. x^4+x^2y^2+y^4


2.(a+2b+c)^3-(a+b)^3-(b+c)^3

3. (x+1)(x+2)+1/4
 
Jun 2009
806
275
How do you factorize the following?

1. x^4+x^2y^2+y^4


2.(a+2b+c)^3-(a+b)^3-(b+c)^3

3. (x+1)(x+2)+1/4

How do you factorize the following?

1. x^4+x^2y^2+y^4


2.(a+2b+c)^3-(a+b)^3-(b+c)^3

3. (x+1)(x+2)+1/4
1) \(\displaystyle x^4 + x^2y^2 + y^4 = x^4 + 2x^2y^2 + y^4 - x^2y^2\)

= \(\displaystyle (x^2 + y^2)^2 - (xy)^2\)

Now simplify.

For (2), use \(\displaystyle x^3 + y^3 = (x+y)(x^2 -xy+y^2)\) Simplify the second and third term. Then proceed.

For (3), simplify the brackets and proceed.
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, mathhomework!

1. Factor: .\(\displaystyle x^4+x^2y^2+y^4\)

Add and subtract \(\displaystyle x^2y^2\!:\)

. . \(\displaystyle x^4 + x^2y^2 {\color{red}\:+\: x^2y^2} + y^4 {\color{red}\:-\: x^2y^2}\)

. . . . . \(\displaystyle =\;(x^4 + 2x^2y^2 + y^4) - x^2y^2\)

. . . . . \(\displaystyle =\; (x^2+ y^2)^2 - (xy)^2 \quad \leftarrow\:\text{ difference of squares}\)

. . . . . \(\displaystyle =\;(x^2+y^2 - xy)(x^2+y^2+xy) \)




3. Factor: .\(\displaystyle (x+1)(x+2)+\tfrac{1}{4}\)

We have: .\(\displaystyle x^2 + 3x + 2 + \tfrac{1}{4} \;=\;x^2 + 3x + \tfrac{9}{4} \;=\;\left(x + \tfrac{3}{2}\right)^2\)



Darn! . . . too slow ... again!
.
 
Last edited:
  • Like
Reactions: mathhomework

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, mathhomework!

I think I've got #2 . . .


\(\displaystyle 2.\;\;(a+2b+c)^3-(a+b)^3-(b+c)^3 \)

\(\displaystyle \text{We have: }\;(a+2b+c)^3 - \underbrace{\bigg[(a+b)^3 + (b+c)^3\bigg]}_{\text{sum of cubes}}\)

. . . . . \(\displaystyle =\;(a+2b+c)^3 - \bigg[(a+b)(b+c)\bigg]\bigg[(a+b)^2 - (a+b)(b+c) + (b+c)^2\bigg] \)

. . . . . \(\displaystyle =\; (a+2b+c)^3 - (a +2b+c)(a^2 + b^2 + c^2 + ab + bc - ac)\)


\(\displaystyle \text{Factor: }\;(a+2b+c)\,\bigg[(a+2b+c)^2 - (a^2+b^2+c^2 + ab + bc - ac)\bigg]\)

. . . . . \(\displaystyle =\;(a+2b+c)\left(3b^2 + 3ab + 3bc + 3ac\right)\)

. . . . . \(\displaystyle =\; 3(a+2b+c)\left(b^2 + ab + bc + ac\right)\)

. . . . . \(\displaystyle =\;3(a+2b+c)\bigg[b(b+a) + c(b + a)\bigg]\)

. . . . . \(\displaystyle =\;3(a+2b+c)(b+a)(b+c)\)


\(\displaystyle \text{Answer: }\;3(a+b)(b+c)(a + 2b + c)\)

 
  • Like
Reactions: mathhomework