Question 766755
There are special products that are probably shown as formulas in your textbook, and are useful in factoring. Some examples:
Square of a binomial:
{{{(A+B)^2=A^2+2AB+B^2}}} and {{{(A-B)^2=A^2-2AB+B^2}}}
Cube of a binomial:
{{{(A+B)^3=A^3+3A^2B+3AB^2+B^3}}} and {{{(A-B)^3=A^3-3A^2B+3AB^2-B^3}}}
Difference of squares:
{{{(A+B)(A-B)=A^2-B^2}}}
Sum and difference of cubes:
{{{(A-B)(A^2+AB+B^2)=A^3-B^3}}} and {{{(A+B)(A^2-AB+B^2)=A^3+B^3}}}
 
There is also taking out a common factor (if there is one), and factoring by parts (if applicable).
 
{{{a^9 - 1 =(a^3)^3-1=((a^3)-1)((a^3)^2(1)+(a^3)1^2+1^3)}}}={{{highlight((a^3-1)(a^6+a^3+1))}}}
A {{{highlight((a^3-1)(a^6+a^3+1))}}}
 
{{{x^8 - 16x^4 =(x^4)(x^4-16)=x^4((x^2)^2-4^2)=x^4(x^2+4)(x^2-4)=x^4(x^2+4)(x^2-2^2)=x^4(x^2+4)(x+2)(x-2)
A {{{highlight(x4(x - 2)(x + 2)(x2 + 4) )}}}
E {{{x^4(x^4 - 16)}}} is a factorization, but is not a {{{highlight(complete)}}} factorization, because you can factor further
F cannot be factored
 
{{{a4 + 3a3+ 27a + 81}}} can be factored by parts:
{{{a4 + 3a3+ 27a + 81=(a4 + 3a3)+ (27a + 81)=a^3(a+3)+27(a+3)=(a+3)(a^3+27)=(a+3)(a^3+3^3)}}}
At this point, you factor {{{a^3+3^3}}} as a special product and get
{{{a4 + 3a3+ 27a + 81=(a+3)(a^3+3^3)=(a+3)(a+3)(a^2+3a+3^2)=(a+3)^2(a^2+3a+3^2)}}}
E {{{highlight(((a + 3)^2)(a2 - 3a + 9))}}}