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Factor completely:
$3a^2 - 8a + 4$
$16a^4 - 1$
$2x^2 + 2xy - 3x - 3y$
$4z^3 - 32$
$x^2 - 2x + 1 - z^2$
$x^3 + 2x^2 - 3x - 6$
$24x + 144 + x^2$
$8x^3 - 1$
$6x^3 + 48$
$9x^2 - 30x + 25$
$9x^2 + 6xy + y^2 - 4$
$x^4 + xy^3 + 4yx^3 + 4y^4$
$2x^4 - 2y^2 - 4y - 2$
$2x^3+4x^2-x-2$
$16b^3-2a^3$
$2x^4 - 6x^3 - 20x^2$
$128x^6 - 2y^6z^{12}$
$x^3 + y^3 + 3y^2 + 3y + 1$
$9x^2 - 6xy + y^2 - 25$
${\displaystyle{(3a-2)(a-2)}}$
${\displaystyle{ \begin{array}[t]{rcl} 16a^4 - 1 &=& (4a^2 - 1)(4a^2 + 1)\\ &=& \fbox{$(2a-1)(2a+1)(4a^2 +1)$} \end{array}}}$
$\displaystyle{ \begin{array}[t]{rcl} 2x^2 + 2xy - 3x - 3y &=& (2x^2 + 2xy) + (-3x - 3y)\\ &=& 2x(x + y) - 3(x + y) \\ &=& \fbox{$(2x - 3)(x + y)$} \end{array}}$
$\displaystyle{ \begin{array}[t]{rcl} 4z^3 - 32 &=& 4(z^3-8)\\ &=& \fbox{$4(z-2)(z^2+2z+4)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} x^2 - 2x + 1 - z^2 &=& (x^2 - 2x + 1) - z^2\\ &=& (x - 1)^2 - z^2\\ &=& \fbox{$(x - 1 + z)(x - 1 - z)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} x^3 + 2x^2 - 3x - 6 &=& (x^3+2x^2) + (-3x-6)\\ &=& x^2(x+2) - 3(x+2)\\ &=& \fbox{$(x^2 - 3)(x+2)$} \end{array} }$
$\displaystyle{ (x+12)^2 }$
$\displaystyle{ (2x-1)(4x^2+2x+1) }$
$\displaystyle{ \begin{array}[t]{rcl} 6x^3 + 48 &=& 6(x^3+8)\\ &=& \fbox{$6(x+2)(x^2-2x+4)$} \end{array} }$
$\displaystyle{ (3x-5)^2 }$
$\displaystyle{ \begin{array}[t]{rcl} 9x^2 + 6xy + y^2 - 4 &=& (9x^2 + 6xy + y^2) - 4\\ &=& (3x+y)^2 - 4\\ &=& \fbox{$(3x+y-2)(3x+y+2)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} x^4 + xy^3 + 4yx^3 + 4y^4 &=& (x^4 + xy^3) + (4yx^3 + 4y^4)\\ &=& x(x^3+y^3) + 4y(x^3+y^3)\\ &=& (x+4y)(x^3+y^3)\\ &=& \fbox{$(x+4y)(x+y)(x^2-xy+y^2)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} 2x^4 - 2y^2 - 4y - 2 &=& 2(x^4-y^2-2y-1)\\ &=& 2(x^4 - (y^2+2y+1))\\ &=& 2(x^4 - (y+1)^2)\\ &=& 2(x^2 + (y+1))(x^2 - (y+1))\\ &=& \fbox{$2(x^2+y+1)(x^2-y-1)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} 2x^3+4x^2-x-2 &=& (2x^3+4x^2) + (-x-2)\\ &=& 2x^2(x+2) - (x+2)\\ &=& \fbox{$(2x^2-1)(x+2)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} 16b^3-2a^3 &=& 2(8b^3-a^3)\\ &=& \fbox{$2(2b - a)(4b^2 +2ab + a^2)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} 2x^4 - 6x^3 - 20x^2 &=& 2x^2(x^2-3x-10)\\ &=& \fbox{$2x^2(x-5)(x+2)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} 128x^6 - 2y^6z^{12} &=& 2(64x^6-y^6 z^12)\\ &=& 2(8x^3-y^3 z^6)(8x^3+y^3 z^6)\\ &=& \fbox{$2(2x-yz^2)(2x+yz^2)(4x^2+2xyz^2+y^2z^4)(4x^2-2xyz^2+y^2z^4)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} x^3 + y^3 + 3y^2 + 3y + 1 &=& x^3 + (y^3 + 3y^2 + 3y + 1)\\ &=& x^3 + (y+1)^3\\ &=& (x + (y+1))(x^2 - x(y+1) + (y+1)^2)\\ &=& \fbox{$(x+y+1)(x^2-xy-x+y^2+2y+1)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} 9x^2 - 6xy + y^2 - 25 &=& (9x^2 - 6xy + y^2) - 25\\ &=& (3x-y)^2 - 25\\ &=& \fbox{$(3x-y-5)(3x-y+5)$} \end{array} }$
Factor completely:
$x^3-x$
$27x^3-1$
$4x^4-13x^2+9$
$3x^2-17x+10$
$3x^3-3xy-2x^2y+2y^2$
$6x^2+7x-3$
$8y^6+125y^3$
$9x^2+6x+1$
$3x^2+18x+15$
$4ax-2bx+2ay-by$
$32a^4c-2b^4c$
$a^6-64$
$2x^4-17x^2-9$
$2x^4-6x^3-20x^2$
$3x^4+375x$
$xy^2+y^2-4x-4$
$\displaystyle{ \begin{array}[t]{rcl} x^3-x &=& x(x^2 - 1)\\ &=& \fbox{$x(x-1)(x+1)$} \end{array} }$
$\displaystyle{ (3x-1)(9x^2+3x+1) }$
$\displaystyle{ \begin{array}[t]{rcl} 4x^4-13x^2+9 &=& (4x^2-9)(x^2-1)\\ &=& \fbox{$(2x+3)(2x-3)(x+1)(x-1)$} \end{array} }$
$\displaystyle{ (x-5)(3x-2) }$
$\displaystyle{ \begin{array}[t]{rcl} 3x^3-3xy-2x^2y+2y^2 &=& (3x^3-3xy)+(-2x^2y+2y^2)\\ &=& 3x(x^2-y)-2y(x^2-y)\\ &=& \fbox{$(3x-2y)(x^2-y)$} \end{array} }$
$\displaystyle{ (2x+3)(3x-1) }$
$\displaystyle{ \begin{array}[t]{rcl} 8y^6+125y^3 &=& y^3(8y^3+125)\\ &=& \fbox{$y^3(2y+5)(4y^2-10y+25)$} \end{array} }$
$\displaystyle{ (3x+1)^2 }$
$\displaystyle{ \begin{array}[t]{rcl} 3x^2+18x+15 &=& 3(x^2+6x+5)\\ &=& \fbox{$3(x+5)(x+1)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} 4ax-2bx+2ay-by &=& (4ax-2bx) + (2ay-by)\\ &=& 2x(2a-b)+y(2a-b)\\ &=& \fbox{$(2x+y)(2a-b)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} 32a^4c-2b^4c &=& 2c(16a^4-b^4)\\ &=& 2c(4a^2-b^2)(4a^2+b^2)\\ &=& \fbox{$2c(2a-b)(2a+b)(4a^2+b^2)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} a^6-64 &=& (a^3-8)(a^3+8)\\ &=& \fbox{$(a-2)(a^2+2a+4)(a+2)(a^2-2a+4)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} 2x^4-17x^2-9 &=& (2x^2+1)(x^2-9)\\ &=& \fbox{$(2x^2+1)(x+3)(x-3)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} 2x^4-6x^3-20x^2 &=& 2x^2(x^2-3x-10)\\ &=& \fbox{$2x^2(x-5)(x+2)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} 3x^4+375x &=& 3x(x^3+125)\\ &=& \fbox{$3x(x+5)(x^2-5x+25)$} \end{array} }$
$\displaystyle{ \begin{array}[t]{rcl} xy^2+y^2-4x-4 &=& (xy^2+y^2) + (-4x-4)\\ &=& y^2(x+1) - 4(x+1)\\ &=& (y^2-4)(x+1)\\ &=& \fbox{$(y-2)(y+2)(x+1)$} \end{array} }$
Use the binomial theorem -- and in particular, the related trick for finding quickly all of the coefficients of $(x+y)^n$ without either calculating each ${}_n C_k$ individually or finding rows in Pascal's Triangle above the one in question -- to expand the following:
$(a+b)^9$
$(2x-1)^5$
$(3x^5+2y^4)^6$
$(x+y+z)^4$
First, we calculate the binomial coefficients needed:
$\displaystyle{ \begin{array}[t]{rcl} 1 \cdot \frac{9}{1} &=& 9\\ 9 \cdot \frac{8}{2} &=& 36\\ 36 \cdot \frac{7}{3} &=& 84\\ 84 \cdot \frac{6}{4} &=& 126\\ 126 \cdot \frac{5}{5} &=& 126 \quad {\scriptstyle \textrm{(by symmetry, at this point the numbers start repeating -- only backwards)}}\\ 126 \cdot \frac{4}{6} &=& 84\\ 84 \cdot \frac{3}{7} &=& 36\\ 36 \cdot \frac{2}{8} &=& 9\\ 9 \cdot \frac{1}{9} &=& 1 \quad {\scriptstyle \textrm{(returning to $1$, we have found the last coefficient needed)}}\\ \end{array}}$Thus, $\displaystyle{ \begin{array}[t]{rcl} (a+b)^9 &=& a^9 + 9a^8b + 36a^7b^2 + 84a^6b^3 + 126a^5b^4 + \\ & & \quad 126a^4b^5 + 84a^3b^6 + 36a^2b^7 + 9ab^8 + b^9 \end{array}}$
We know from the sequence of calculations $1 \cdot \frac{5}{1} = 5$, $5 \cdot \frac{4}{2} = 10$, and $10 \cdot \frac{3}{3} = 10$, along with the expected symmetry seen in the rows of Pascal's triangle that $$(a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$$
In attempting to expand $(2x-1)^5$ observe that $a=2x$ and $b=-1$. Substituting these into the expression above, we discover $$(2x-1)^5 = 32x^5 -80x^4 + 80x^3 -40x^2 +10x - 1$$
We know from the sequence of calculations $1 \cdot \frac{6}{1} = 6$, $6 \cdot \frac{5}{2} = 15$, $15 \cdot \frac{4}{3} = 20$, along with the expected symmetry seen in the rows of Pascal's triangle that $$(a+b)^6 = a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6$$
In attempting to expand $(3x^5+2y^4)^6$ observe that $a=3x^5$ and $b=2y^4$. Substituting these into the expression above, we discover $$\begin{array}{rcl} (3x^5+2y^4)^6 &=& 729x^{30} + 2916x^{25}y^4 + 4860 x^{20}y^8 + 4320 x^{15}y^{12} +\\ & & \quad 2160 x^{10} y^{16} + 576 x^5 y^{20} + 64 y^{24} \end{array}$$
Note that we can group the last two terms inside the parentheses so that we can view this as a power of a binomial: $$\begin{array}{rcl} (x+y+z)^4 &=& (x+(y+z))^4\\ &=& x^4 + 4x^3(y+z) + 6x^2(y+z)^2 + 4x(y+z)^3 + (y+z)^4\\ &=& x^4 + 4x^3y + 4x^3z + 6x^2(y^2 + 2yz + z^2) + {}\\ & & \quad 4x(y^3 + 3y^2z + 3yz^2 + z^3) + (y^4 + 4y^3z + 6y^2z^2 + 4yz^3 + z^4)\\ &=& x^4 + 4x^3y + 4x^3z + 6x^2y^2 + 12x^2yz + 6x^2z^2 + {}\\ & & \quad 4xy^3 + 12xy^2z + 12xyz^2 + 4xz^3 + y^4 + 4y^3z + 6y^2z^2 + 4yz^3 + z^4 \end{array}$$
For each, decide if Eisenstein's Criterion can be used to prove the given polynomial is irreducible. If it can, find the prime $p$ in question.
$x^3 + 6x^2 + 9x + 3$
$x^3 - 2x + 50$
$x^3 + 6x^2 - x - 6$
$x^3 + 7x^2 - 2x - 1$
$x^4 + x^3 + x^2 + x + 1$
Eisenstein applies ($p=3$). So $x^3 + 6x^2 + 9x + 3$ is irreducible and thus does not factor.
Eisenstein applies ($p=2$). So $x^3 - 2x + 50$ is irreducible and thus does not factor.
Eisenstein does not apply. Indeed, $x^3 + 6x^2 - x - 6$ is definitely not irreducible as it factors into $(x+1)(x-1)(x+6)$
Eisenstein doesn't apply directly, as no prime goes evenly into the last term of $-1$. Interestingly, note that a polynomial $f(x)$ factors if and only if $f(x+h)$ factors. If only we could find an $h$ where Eisenstein could apply here! Note that $f(x+1) = x^3 + 10x^2 + 15x + 5$, where Eisenstein applies with $p=5$. As $f(x+1)$ is then irreducible, so must $f(x)$ be irreducible!
Eisenstein doesn't apply directly, as no primes divide $1$. That said, we can pull the same trick suggested by the previous problem. If $f(x) = x^4 + x^3 + x^2 + x + 1$, then $f(x+1) = x^4 + 5x^3 + 10x^2 + 10x + 5$. Eisenstein does apply to this new polynomial with $p=5$ and thus proves the irreducibility of the original $f(x)$ polynomial.
Factor $a^4 + 4b^4$
Add a well-chosen value of zero like the one below: $$2a^3 b - 2a^3 b + 2a^2 b^2 + 2a^2b^2 - 4a^2b^2 + 4ab^3 -4ab^3$$ Then experiment with $3$ groups of $3$ terms each.