# Factor Equations

Question 1. Factor $(x+1)^4+(x+3)^4-272$.

Solution: $$\begin{eqnarray}&=&(x+1)^4+(x+3)^4-272\\&=&(x+1)^4+(x+3)^4-272+16-16\\ &=&(x+1)^4+(x+3)^4-256-16\\ &=&\left[(x+1)^4-16\right]+\left[(x+3)^4-256\right]\\ &=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\right]+\left[(x+3)^2+16\right]\left[(x+3)^2-16\right]\\ &=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\right]+\left[(x+3)^2+16\right]\left[(x+3)-4\right]\left[(x+3)+4\right]\end{eqnarray}.$$

Question 2. Factor $x^4+(x+y)^4+y^4$

Solution: $$\begin{eqnarray}&=&(x^4+y^4)+(x+y)^4\\ &=&(x^4+y^4)+(x+y)^4+2x^2y^2-2x^2y^2\\ &=&(x^4+2x^2y^2+y^4)+(x+y)^4-2x^2y^2\\ &=&(x^2+y^2)^2+(x+y)^4-2x^2y^2 \end{eqnarray}$$

I am stuck in question number 2, I dont know what is next after that line.

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$$\begin{split} \ & x^4+y^4+(x+y)^4\\ \ =& (x^2+y^2)^2-2x^2y^2+(x^2+y^2+2xy)^2\\ \ =& (x^2+y^2)^2-2x^2y^2+(x^2+y^2)^2+4xy(x^2+y^2)+4x^2y^2\\ \ =& 2((x^2+y^2)^2+x^2y^2+2xy(x^2+y^2))\\ \ =& 2(x^2+y^2+xy)^2 \end{split}$$

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\begin{align*} (x+y)^4+x^4+y^4&=2(x^4+2x^3y+3x^2y^2+2xy^3+y^4)\\ &=2(x^4+2x^3y+2x^2y^2+x^2 y^2+2 x y^3+y^4)\\ &=2(x^4+2(xy+y^2)x^2+(xy+y^2)^2)\\ &=2(x^2+xy+y^2)^2 \end{align*}

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You are very welcome! – BlackAdder Jul 7 '13 at 11:00

For the first, I will put $y=\frac{x+1+x+3}2=x+2$

so that $x +1=y-1, x+3=y+1$ and the odd powers of $y$ vanish in $(y-1)^4+(y+1)^4$

$$\implies (x+1)^4+(x+3)^4-272=(y-1)^4+(y+1)^4-272$$

$$=2\{y^4+6y^2+1\}-272=2(y^4+6y^2-135)$$

$$=2\{y^4+(15-9)y^2-135\}=2(y^2+15)(y^2-9) =2(y^2-15)(y+3)(y-3)$$

$$=2\{(x+2)^2-15\}(x+5)(x-1)$$

$$\text{Now, }(x+2)^2-15=x^2+4x+4-15=x^2+4x-11\text{ which is not reducible}$$

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Thank you lab bhattacharjee! – Al-Ahmadgaid Asaad Jul 7 '13 at 10:58