Factorization

**mathhomework** · May 28th 2010, 01:24 AM

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

**sa-ri-ga-ma** · May 28th 2010, 03:10 AM

Originally Posted by mathhomework

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

Originally Posted by mathhomework

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) $x^4 + x^2y^2 + y^4 = x^4 + 2x^2y^2 + y^4 - x^2y^2$

= $(x^2 + y^2)^2 - (xy)^2$

Now simplify.

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

For (3), simplify the brackets and proceed.

**Soroban** · May 28th 2010, 03:20 AM

Hello, mathhomework!

1. Factor: . $x^4+x^2y^2+y^4$

Add and subtract $x^2y^2\!:$

. . $x^4 + x^2y^2 {\color{red}\:+\: x^2y^2} + y^4 {\color{red}\:-\: x^2y^2}$

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

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

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

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

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

Darn! . . . too slow ... again!
.

**Soroban** · May 28th 2010, 03:57 AM

Hello, mathhomework!

I think I've got #2 . . .

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

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

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

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

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

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

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

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

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

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

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