factorize p3- 27 + 3p - 9p
Answers
hi here is ur answer!!
Step by Step Solution:
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "p2" was replaced by "p^2". 1 more similar replacement(s).
STEP
1
:
Equation at the end of step 1
(((p3) + 3p2) - 9p) - 27
STEP
2
:
Checking for a perfect cube
2.1 p3+3p2-9p-27 is not a perfect cube
Trying to factor by pulling out :
2.2 Factoring: p3+3p2-9p-27
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: -9p-27
Group 2: 3p2+p3
Pull out from each group separately :
Group 1: (p+3) • (-9)
Group 2: (p+3) • (p2)
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Add up the two groups :
(p+3) • (p2-9)
Which is the desired factorization
Trying to factor as a Difference of Squares:
2.3 Factoring: p2-9
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : 9 is the square of 3
Check : p2 is the square of p1
Factorization is : (p + 3) • (p - 3)
Multiplying Exponential Expressions:
2.4 Multiply (p + 3) by (p + 3)
The rule says : To multiply exponential expressions which have the same base, add up their exponents.
In our case, the common base is (p+3) and the exponents are :
1 , as (p+3) is the same number as (p+3)1
and 1 , as (p+3) is the same number as (p+3)1
The product is therefore, (p+3)(1+1) = (p+3)2
Final result :
(p + 3)2 • (p - 3)