1
$\begingroup$

Please check my answer in factoring this 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.

$\endgroup$
1
$\begingroup$

\begin{equation} \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} \end{equation}

$\endgroup$
  • $\begingroup$ Thank you Samrat Mukhopadhyay! $\endgroup$ – Al-Ahmadgaid Asaad Jul 7 '13 at 10:57
  • 1
    $\begingroup$ welcome! @al-ahmadgaid-asaad $\endgroup$ – Samrat Mukhopadhyay Jul 7 '13 at 11:08
1
$\begingroup$

\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*}

$\endgroup$
  • 1
    $\begingroup$ You are very welcome! $\endgroup$ – BlackAdder Jul 7 '13 at 11:00
1
$\begingroup$

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 $(y-1)^4+(y+1)^4$

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

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

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

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

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

$\endgroup$
  • $\begingroup$ Thank you lab bhattacharjee! $\endgroup$ – Al-Ahmadgaid Asaad Jul 7 '13 at 10:58
  • 1
    $\begingroup$ @Al-AhmadgaidAsaad, my pleasure $\endgroup$ – lab bhattacharjee Jul 7 '13 at 11:39
1
$\begingroup$

Let $w = x + 2$

\begin{align} (x+1)^4+(x+3)^4-272 &= (w-1)^4+(w+1)^4-272\\ &= 2w^4 + 12w^2 - 270\\ &= 2(w^4 + 6w - 135)\\ &= 2(w^4 +6w^2 + 9 - 144)\\ &= 2(w^2 + 3)^2 - 12^2)\\ &= 2(w^2 + 15)(w^2 - 9)\\ &= 2(w^2 + 15)(w - 3)(w + 3)\\ &=2(x^2 + 4x + 19)(x - 1)(x + 5)\\ \end{align}

\begin{align} (x^4+y^4)+(x+y)^4 &= 2x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + 2y^4\\ &= 2(x^4 + 2x^3y + 3x^2y^2 + 2xy^3 + y^4)\\ &= 2((x^4 + x^3y + x^2y^2) + (x^3y + x^2y^2 + xy^3) + (x^2y^2 + xy^3 + y^4))\\ &= 2(x^2(x^2 + xy + y^2) + xy(x^2 + xy + y^2) + y^2(x^2 + xy + y^2))\\ &= 2(x^2 + xy + y^2)^2 \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.