# Factorization

#### 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

#### sa-ri-ga-ma

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
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!
.

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mathhomework

#### Soroban

MHF Hall of Honor
Hello, mathhomework!

I think I've got #2 . . .

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

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

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

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

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

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

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

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

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

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

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