google "sum of cubes"
x^3 + 2^3
Factor the polynomial
X^3 + 8
2 answers
To factor x³ + 8 you must use sum of cubes formula:
A³ + B³ = ( A + B ) ( A² - A ∙ B + B² )
In this case :
A = x
B = 2
becouse 2³ = 8
So:
x³ + 8 = x³ + 2³
x³ + 2³ = ( x + 2 ) ( x² - x ∙ 2 + 2² ) =
( x + 2 ) ( x² - 2 x + 4 )
x² - 2 x + 4 can't be factoring becouse discriminant of x² - 2 x + 4:
D = b² - 4 ∙ c = ( - 2 )² - 4 ∙ 1 ∙ 4 = 4 - 16 = - 12
This mean:
x² - 2 x + 4 has two complex solutions and can't be factoring with real numbers.
Solution:
x³ + 8 = ( x + 2 ) ( x² - 2 x + 4 )
A³ + B³ = ( A + B ) ( A² - A ∙ B + B² )
In this case :
A = x
B = 2
becouse 2³ = 8
So:
x³ + 8 = x³ + 2³
x³ + 2³ = ( x + 2 ) ( x² - x ∙ 2 + 2² ) =
( x + 2 ) ( x² - 2 x + 4 )
x² - 2 x + 4 can't be factoring becouse discriminant of x² - 2 x + 4:
D = b² - 4 ∙ c = ( - 2 )² - 4 ∙ 1 ∙ 4 = 4 - 16 = - 12
This mean:
x² - 2 x + 4 has two complex solutions and can't be factoring with real numbers.
Solution:
x³ + 8 = ( x + 2 ) ( x² - 2 x + 4 )