How many factors are possible of 27x^3 - 1?
Answers
Answer:
(3x - 1) • (9x2 + 3x + 1)
Look down for explanation ^-^
Step-by-step explanation:
STEP 1 :
Equation at the end of step 1
33x3 - 1
STEP 2 :
Trying to factor as a Difference of Cubes
2.1 Factoring: 27x3-1
Theory : A difference of two perfect cubes, a3 - b3 can 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+b3
Check : 27 is the cube of 3
Check : 1 is the cube of 1
Check : x3 is the cube of x1
Factorization is :
(3x - 1) • (9x2 + 3x + 1)
Trying to factor by splitting the middle term
2.2 Factoring 9x2 + 3x + 1
The first term is, 9x2 its coefficient is 9 .
The middle term is, +3x its coefficient is 3 .
The last term, "the constant", is +1
Step-1 : Multiply the coefficient of the first term by the constant 9 • 1 = 9
Step-2 : Find two factors of 9 whose sum equals the coefficient of the middle term, which is 3 .
-9 + -1 = -10
-3 + -3 = -6
-1 + -9 = -10
1 + 9 = 10
3 + 3 = 6
9 + 1 = 10
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Final result :
(3x - 1) • (9x2 + 3x + 1)