Factoring Polynomials: Very Difficult Problems with Solutions

Solution:
[tex]3x^{3} - x^{2}y + 6x^{2}y - 2xy^{2} + 3xy^{2} - y^{3}=[/tex]
[tex]x^{2}(3x - y) + 2xy(3x - y) + y^{2}(3x-y) =[/tex]
[tex](3x - y)(x^{2} + 2xy + y^{2})=[/tex]
[tex](3x - y)(x + y)^{2}[/tex]

Solution:
[tex](x + 4)^{3} - 9x - 36=[/tex]
[tex](x + 4)^{3} - 9(x + 4)=[/tex]
[tex](x + 4)[(x + 4)^{2} - 9]=[/tex]
[tex](x + 4)(x + 4 - 3)(x + 4 + 3)=[/tex]
[tex](x + 4)(x + 1)(x + 7)[/tex]

Solution:
[tex]x^{2} + 2x + 1 - y^{2}=[/tex]
[tex](x^{2} + 2x + 1) - y^{2}=[/tex]
[tex](x + 1)^{2} - y^{2}=[/tex]
[tex](x + 1 - y)(x + 1 + y)[/tex]

Solution:
[tex]x^{2} - 6x + 9 - 4y^{2}=[/tex]
[tex](x^{2} - 6x + 9) - 4y^{2}=[/tex]
[tex](x - 3)^{2} - (2y)^{2}=[/tex]
[tex](x - 3 - 2y)(x - 3 + 2y)[/tex]

Solution:
[tex]a^{2} - b^{2} - 10b - 25 =[/tex]
[tex]a^{2} - (b^{2} + 10b + 25) =[/tex]
[tex]a^{2} - (b + 5)^{2}=[/tex]
[tex][a - (b + 5)][a + (b + 5)] =[/tex]
[tex](a - b - 5)(a + b + 5)[/tex]

Solution:
[tex]a^{2} - b^{2} + 12b - 36 =[/tex]
[tex]a^{2} - (b^{2} - 12b + 36)=[/tex]
[tex]a^{2} - (b - 6)^{2}=[/tex]
[tex][a - (b - 6)][a + (b - 6)]=[/tex]
[tex](a - b + 6)(a + b - 6)[/tex]

Solution:
Add and subtract [tex]4x^2[/tex] to get a perfect square:
[tex]x^4+4=x^4+4x^2+4-4x^2=(x^2+2)^2-(2x)^2[/tex]
[tex]=(x^2+2-2x)(x^2+2+2x)=(x^2-2x+2)(x^2+2x+2)[/tex]

Solution:
Subtract and add [tex]x^2[/tex]:
[tex]x^5+x+1=(x^5-x^2)+(x^2+x+1)=x^2(x^3-1)+(x^2+x+1)[/tex]
[tex]=x^2(x-1)(x^2+x+1)+(x^2+x+1)=(x^2+x+1)\bigl(x^2(x-1)+1\bigr)[/tex]
[tex]=(x^2+x+1)(x^3-x^2+1)[/tex]

Solution:
Group as a difference of cubes and a sum of cubes:
[tex]\bigl((x+y+z)^3-x^3\bigr)-(y^3+z^3)[/tex]
[tex]=(y+z)\bigl((x+y+z)^2+(x+y+z)x+x^2\bigr)-(y+z)(y^2-yz+z^2)[/tex]
After expanding the big bracket, [tex]y^2[/tex] and [tex]z^2[/tex] cancel:
[tex]=(y+z)(3x^2+3xy+3yz+3zx)=3(y+z)\bigl(x(x+y)+z(x+y)\bigr)=3(x+y)(y+z)(z+x)[/tex]

Solution:
Since [tex]c-a=-(a-b)-(b-c)[/tex],
[tex]a^3(b-c)+b^3(c-a)+c^3(a-b)=a^3(b-c)-b^3(b-c)-b^3(a-b)+c^3(a-b)[/tex]
[tex]=(b-c)(a^3-b^3)-(a-b)(b^3-c^3)[/tex]
[tex]=(a-b)(b-c)(a^2+ab+b^2)-(a-b)(b-c)(b^2+bc+c^2)[/tex]
[tex]=(a-b)(b-c)(a^2-c^2+ab-bc)[/tex]
[tex]a^2-c^2+ab-bc=(a-c)(a+c)+b(a-c)=(a-c)(a+b+c)[/tex]
Therefore [tex]a^3(b-c)+b^3(c-a)+c^3(a-b)=(a-b)(b-c)(a-c)(a+b+c)[/tex]

Solution:
Write [tex]2a^2b^2=4a^2b^2-2a^2b^2[/tex]:
[tex]4a^2b^2-(a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2)=(2ab)^2-(a^2+b^2-c^2)^2[/tex]
[tex]=(2ab-a^2-b^2+c^2)(2ab+a^2+b^2-c^2)=\bigl(c^2-(a-b)^2\bigr)\bigl((a+b)^2-c^2\bigr)[/tex]
[tex]=(c-a+b)(c+a-b)(a+b-c)(a+b+c)[/tex]

Solution:
Let [tex]x=a-b[/tex], [tex]y=b-c[/tex], [tex]z=c-a[/tex]. Then [tex]x+y+z=0[/tex], so [tex]z=-(x+y)[/tex] and
[tex]x^3+y^3+z^3=x^3+y^3-(x+y)^3=-3x^2y-3xy^2=-3xy(x+y)=3xyz[/tex]
Therefore [tex](a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)[/tex], [tex]k=3[/tex].
Check: [tex]a=0[/tex], [tex]b=1[/tex], [tex]c=2[/tex]: [tex](-1)^3+(-1)^3+2^3=6=3\times (-1)\times (-1)\times 2[/tex]