Step 1 :

Trying to factor as a distinction of Squares:1.1 Factoring: x6-64 theory : A difference of 2 perfect squares, A2-B2can it is in factored into (A+B)•(A-B)Proof:(A+B)•(A-B)= A2 - AB+BA-B2= A2 -AB+ abdominal - B2 = A2 - B2Note : abdominal muscle = BA is the commutative property of multiplication. Note : -AB+ abdominal equates to zero and also is as such eliminated native the expression.Check: 64 is the square that 8Check: x6 is the square of x3Factorization is :(x3 + 8)•(x3 - 8)

Trying to factor as a amount of Cubes:

1.2 Factoring: x3 + 8 Theory:A amount of two perfect cubes, a3+b3 have the right to be factored into :(a+b)•(a2-ab+b2)Proof: (a+b)•(a2-ab+b2) = a3-a2b+ab2+ba2-b2a+b3=a3+(a2b-ba2)+(ab2-b2a)+b3=a3+0+0+b3=a3+b3Check:8is the cube the 2Check: x3 is the cube of x1Factorization is :(x + 2)•(x2 - 2x + 4)

Trying to variable by separating the center term

1.3Factoring x2 - 2x + 4 The very first term is, x2 the coefficient is 1.The middle term is, -2x its coefficient is -2.The critical term, "the constant", is +4Step-1 : multiply the coefficient of the an initial term by the continuous 1•4=4Step-2 : uncover two components of 4 who sum equates to the coefficient that the middle term, i beg your pardon is -2.

-4+-1=-5
-2+-2=-4
-1+-4=-5
1+4=5
2+2=4
4+1=5

Observation : No two such components can be uncovered !! Conclusion : Trinomial deserve to not it is in factored

Trying to variable as a difference of Cubes:1.4 Factoring: x3-8 concept : A difference of two perfect cubes, a3-b3 have the right to be factored into(a-b)•(a2+ab+b2)Proof:(a-b)•(a2+ab+b2)=a3+a2b+ab2-ba2-b2a-b3 =a3+(a2b-ba2)+(ab2-b2a)-b3=a3+0+0-b3=a3-b3Check:8is the cube the 2Check: x3 is the cube the x1Factorization is :(x - 2)•(x2 + 2x + 4)

Trying to variable by separating the middle term

1.5Factoring x2 + 2x + 4 The very first term is, x2 that is coefficient is 1.The center term is, +2x that coefficient is 2.The critical term, "the constant", is +4Step-1 : multiply the coefficient that the first term by the consistent 1•4=4Step-2 : find two factors of 4 whose sum equals the coefficient that the middle term, i m sorry is 2.

-4+-1=-5
-2+-2=-4
-1+-4=-5
1+4=5
2+2=4
4+1=5

Observation : No 2 such determinants can be discovered !! Conclusion : Trinomial can not be factored