## Step 1 :

Trying to factor as a distinction of Squares:1.1 Factoring: x6-64 theory : A difference of two perfect squares, A2-B2can it is in factored into (A+B)•(A-B)Proof:(A+B)•(A-B)= A2 - AB+BA-B2= A2 -AB+ ab - B2 = A2 - B2Note : abdominal = BA is the commutative building of multiplication. Keep in mind : -AB+ ab amounts to zero and also is therefore eliminated from the expression.Check: 64 is the square that 8Check: x6 is the square the x3Factorization is :(x3 + 8)•(x3 - 8)

Trying to element as a sum of Cubes:

1.2 Factoring: x3 + 8 Theory:A amount of 2 perfect cubes, a3+b3 deserve 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 that 2Check: x3 is the cube that x1Factorization is :(x + 2)•(x2 - 2x + 4)

Trying to aspect by splitting the middle term

1.3Factoring x2 - 2x + 4 The first term is, x2 the coefficient is 1.The middle term is, -2x the coefficient is -2.The last term, "the constant", is +4Step-1 : main point the coefficient the the an initial term by the constant 1•4=4Step-2 : discover two factors of 4 whose sum equals the coefficient of 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 2 such determinants can be discovered !! Conclusion : Trinomial have the right to not it is in factored

Trying to element as a distinction of Cubes:1.4 Factoring: x3-8 theory : A distinction of 2 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 of 2Check: x3 is the cube the x1Factorization is :(x - 2)•(x2 + 2x + 4)

Trying to variable by dividing the middle term

1.5Factoring x2 + 2x + 4 The first term is, x2 that coefficient is 1.The middle term is, +2x its coefficient is 2.The critical term, "the constant", is +4Step-1 : multiply the coefficient that the an initial term by the constant 1•4=4Step-2 : find two components of 4 whose sum equates to the coefficient the 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 2 such determinants can be found !! Conclusion : Trinomial can not be factored