Factoring Polynomials: Problems with Solutions

Solution:
[tex]xy + 2x + y + 2 =[/tex]
[tex]x \times y + x \times 2 + 1 \times y + 1 \times 2 =[/tex]
[tex]x \times (y + 2) + 1 \times (y + 2) =[/tex]
[tex](y + 2)(x + 1)[/tex]

Solution:
[tex]3x^{3}+5x^{2}-6x=[/tex]
[tex]x \times 3x^{2}+x \times 5x+x \times (-6) =[/tex]
[tex]x \times (3x^{2}+5x-6)[/tex]

Solution:
[tex]xy - 3x + y - 3 =[/tex]
[tex]x \times y + x \times (-3) + 1 \times y + 1 \times (-3) =[/tex]
[tex]x(y - 3) + 1(y - 3) = (y - 3)(x + 1)[/tex]

Solution:
[tex]xy + 4x + 2y + 8=[/tex]
[tex]x \times y + x \times 4 + 2 \times y + 2 \times 4 =[/tex]
[tex]x(y + 4) + 2(y + 4)=(y + 4)(x + 2)[/tex]

Solution:
$xy-5x+3y-15=$
$x\cdot y+x\cdot(-5)+3\cdot y+3\cdot(-5)=$
$x(y-5)+3(y-5)=(y-5)(x+3)$

Solution:
$xy+7+y+7x=$
$xy+7x+y+7=$
$x\cdot y+x\cdot7+1\cdot y+1\cdot7=$
$x(y+7)+1(y+7)=(y+7)(x+1)$

Solution:
$xy-8+y-8x=$
$xy-8x+y-8=$
$x\cdot y+x\cdot(-8)+1\cdot y+1\cdot(-8)=$
$x(y-8)+1(y-8)=(y-8)(x+1)$

Solution:
$xy+9+3y+3x=$
$xy+3x+3y+9=$
$x\cdot y+x\cdot3+3\cdot y+3\cdot3=$
$x(y+3)+3(y+3)=(y+3)(x+3)$

Solution:
$xy-21+3y-7x=$
$xy-7x+3y-21=$
$x\cdot y+x\cdot(-7)+3\cdot y+3\cdot(-7)=$
$x(y-7)+3(y-7)=(y-7)(x+3)$

Solution:
$xy-x-y+1=$
$x\cdot y +x\cdot(-1)-1\cdot y-1\cdot(-1)=$
$x(y-1)-1(y-1)=(y-1)(x-1) $

Solution:
$xy-4x-y+4=$
$x\cdot y +x\cdot(-4)-1\cdot y-1\cdot(-4)=$
$x(y-4)-1(y-4)=(y-4)(x-1) $

Solution:
$xy+3x-y-3=$
$x\cdot y +x\cdot3-1\cdot y-1\cdot3=$
$x(y+3)-1(y+3)=(y+3)(x-1) $

Solution:
$xy-9x-3y+27=$
$x\cdot y +x\cdot(-9)-3\cdot y-3\cdot(-9)=$
$x(y-9)-3(y-9)=(y-9)(x-3) $

Solution:
[tex]xy + 4x - 5y - 20 =[/tex]
[tex]x \times y + x \times 4 - 5 \times y - 5 \times 4 =[/tex]
[tex]x(y + 4) - 5(y + 4) = (y + 4)(x - 5)[/tex]

Solution:
[tex]2xy + 6x - y - 3=[/tex]
[tex]2x \times y + 2x \times 3 - 1 \times y - 1 \times 3=[/tex]
[tex]2x(y + 3) - 1(y + 3)=(y + 3)(2x - 1)[/tex]

Solution:
$x^{3}+x^{2}+x+1=$
$x^{2}\cdot x+x^{2}\cdot1+1\cdot x+1\cdot1=$
$x^{2}(x+1)+1(x+1)=(x+1)(x^{2}+1)$

Solution:
$x^{4}+4x^{2}+x^{2}+4 =$
$x^{2}\cdot x^{2} + x^{2}\cdot4 + 1\cdot x^{2}+ 1\cdot4 =$
$x^{2}(x^{2}+4)+1(x^{2}+4)=$
$(x^{2}+4)(x^{2}+1)$

Solution:
$x^{4}-3x^{3}+x-3 = $
$x^{3}\cdot x + x^{3}\cdot(-3) +1\cdot x+1\cdot(-3) =$
$x^{3}(x-3)+1(x-3)= (x-3)(x^{3}+1)$

Solution:
$2x^{5} + 2x^{3} + 3x^{2} +3 = $
$2x^{3}\cdot x^{2} + 2x^{2}\cdot1 +3\cdot x^{2}+3\cdot1 = $
$2x^{3}(x^{2}+1)+3(x^{2}+1)= $
$(x^{2}+1)(2x^{3}+3)$

Solution:
$4x^{6}-2x^{5}-6x+3 =$
$ 2x^{5}\cdot 2x + 2x^{5}\cdot(-1) - 3\cdot 2x-3\cdot(-1) =$
$ 2x^{5}(2x-1)-3(2x-1)=$
$ (2x-1)(2x^{5}-3)$

Solution:
$x^{4}+3x^{2}-5x^{2}-15 = $
$x^{2}\cdot x^{2} + x^{2}\cdot3 - 5\cdot x^{2}- 5\cdot3 = $
$x^{2}(x^{2}+3)-5(x^{2}+3)= $
$(x^{2}+3)(x^{2}-5)$

Solution:
$4x^{5}-6x^{3}+6x^{2}-9 =$
$ 2x^{3}\cdot 2x^{2} + 2x^{3}\cdot(-3) +3\cdot x^{2}+3\cdot(-3) =$
$ 2x^{3}(2x^{2}-3)+3(2x^{2}-3)=$
$ (2x^{2}-3)(2x^{3}+3)$

Solution:
$3x^{6}+5x^{4}-6x^{2}-10 =$
$ x^{4}\cdot 3x^{2} + x^{2}\cdot 5x^{2} - 2\cdot 3x^{2}- 2\cdot5 =$
$ x^{4}(3x^{2}+5)-2(3x^{2}+5)=$
$ (3x^{2}+5)(x^{4}-2)$

Solution:
$9x^{4}-6x^{3}-3x+2 =$
$ 3x^{3}\cdot 3x + 3x^{3}\cdot(-2) - 1\cdot 3x-1\cdot(-2) =$
$ 3x^{3}(3x-2)-1(3x-2)= $
$(3x-2)(3x^{3}-1)$

Solution:
We use the formula [tex](a-b)(a+b)=a^2-b^2[/tex]:
[tex](4x-3)^2-(3x+2)^2=\bigl((4x-3)-(3x+2)\bigr)\bigl((4x-3)+(3x+2)\bigr)[/tex]
[tex]=(4x-3-3x-2)(4x-3+3x+2)=(x-5)(7x-1)[/tex]

Solution:
This is the cube of a sum [tex](a+b)^3=a^3+3a^2b+3ab^2+b^3[/tex] with [tex]a=2x[/tex], [tex]b=3[/tex]:
[tex]8x^3+36x^2+54x+27=(2x)^3+3\times (2x)^2\times 3+3\times 2x\times 3^2+3^3=(2x+3)^3[/tex]

Solution:
We use the formula [tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]:
[tex]125x^3-y^3=(5x)^3-y^3=(5x-y)\bigl((5x)^2+5x\times y+y^2\bigr)=(5x-y)(25x^2+5xy+y^2)[/tex]

Solution:
[tex]ab^3-4ab^2+4ab=ab(b^2-4b+4)=ab(b-2)^2[/tex]

Solution:
Since [tex]x^2=x\times x[/tex] and [tex]4=2^2[/tex], the square must be [tex](x+2)^2[/tex]:
[tex](x+2)^2=x^2+2\times x\times 2+2^2=x^2+4x+4 \Rightarrow a=4[/tex]

Solution:
Since [tex]13=2\times 6.5[/tex], this is the square of a sum:
[tex](6.5)^2+2\times 6.5\times 4.5+(4.5)^2=(6.5+4.5)^2=11^2=121[/tex]

Solution:
We use the formula [tex](a-b)(a+b)=a^2-b^2[/tex]:
[tex]779^2-221^2=(779-221)(779+221)=558\times 1000=558000[/tex]

Solution:
[tex]12x-1-36x^2=-(36x^2-12x+1)=-(6x-1)^2\leq 0[/tex]
The value [tex]0[/tex] is reached for [tex]x=\frac{1}{6}[/tex], so the values are not always negative: they are non-positive.

Solution:
[tex]4x^3-x(x+1)^2=x\bigl(4x^2-(x+1)^2\bigr)=x\bigl((2x)^2-(x+1)^2\bigr)[/tex]
[tex]=x(2x-x-1)(2x+x+1)=x(x-1)(3x+1)[/tex]

Solution:
[tex]a^2-b^2=(a-b)(a+b)=5(a+b)=175 \Rightarrow a+b=35[/tex]
[tex]a+b=35[/tex], [tex]a-b=5 \Rightarrow 2a=40 \Rightarrow a=20, b=15[/tex]
Check: [tex]20^2-15^2=400-225=175[/tex]