142 algebra De mystif ieD
EXAMPLES
Completely factor the expression.
4 + 6x
Each term is divisible by 2, so factor 2 from 4 and 6x: 4 + 6x = 2 ⋅ 2 + 2 ⋅ 3 x =
2(2 + 3x).
2 x + 5x^2 = x ⋅ 2 + x ⋅ 5 x = x(2 + 5x)
8 x + 8 = 8 ⋅ x + 8 ⋅ 1 = 8(x + 1)
4 xy + 6x^2 + 2xy^2 = 2x ⋅ 2 y + 2x ⋅ 3 x + 2x ⋅ y^2 = 2x(2y + 3x + y^2 )
3 x^2 + 6x = 3x ⋅ x + 3x ⋅ 2 = 3x(x + 2)
Complicated expressions can be factored in several steps. Take for example
48 x^5 y^3 z^6 + 60x^4 yz^3 + 36x^6 y^2 z. We can begin by dividing 12xyz from each
term, to obtain:
48 x^5 y^3 z^6 + 60x^4 yz^3 + 36x^6 y^2 z = 12xyz ⋅ 4 x^4 y^2 z^5 + 12xyz ⋅ 5 x^3 z^2 + 12xyz ⋅ 3 x^5 y
= 12xyz(4x^4 y^2 z^5 + 5x^3 z^2 + 3x^5 y)
Each term in the parentheses is divisible by x^2 , so it is not completely fac-
tored. We now factor x^2 from each term, and then we will simplify it, thus
obtaining:
4 x^4 y^2 z^5 + 5x^3 z^2 + 3x^5 y = x^2 ⋅ 4 x^2 y^2 z^5 + x^2 ⋅ 5 xz^2 + x^2 ⋅ 3 x^3 y = x^2 (4x^2 y^2 z^5 + 5xz^2 + 3x^3 y)
We now factor the original problem, and obtain:
48 x^5 y^3 z^6 + 60x^4 yz^3 + 36x^6 y^2 z = 12xyz ⋅ x^2 (4x^2 y^2 z^5 + 5xz^2 + 3x^3 y)
= 12x^3 yz(4x^2 y^2 z^5 + 5xz^2 + 3x^3 y)PRACTICE
Completely factor the expression.- 4x − 10y =
- 3x + 6y − 12 =
- 5x^2 + 15 =
- 4x^2 + 4x =
- 4x^3 − 6x^2 + 12x =
- −24xy^2 + 6x^2 + 18x =
- 30x^4 − 6x^2 =
- 15x^3 y^2 z^7 − 30xy^2 z^4 + 6x^4 y^2 z^6 =
EXAMPLES
Completely factor the expression.PRACTICE
Completely factor the expression.