Answer:
STEP
1
:
Equation at the end of step 1
(2x • (y3)) - (2x3 • y)
2
Equation at the end of step
2xy3 - 2x3y
3
4
Pulling out like terms
4.1 Pull out like factors :
2xy3 - 2x3y = -2xy • (x2 - y2)
Trying to factor as a Difference of Squares:
4.2 Factoring: x2 - y2
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 : x2 is the square of x1
Check : y2 is the square of y1
Factorization is : (x + y) • (x - y)
Final result :
-2xy • (x + y) • (x - y)
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Answers & Comments
Answer:
STEP
1
:
Equation at the end of step 1
(2x • (y3)) - (2x3 • y)
STEP
2
:
Equation at the end of step
2
:
2xy3 - 2x3y
STEP
3
:
STEP
4
:
Pulling out like terms
4.1 Pull out like factors :
2xy3 - 2x3y = -2xy • (x2 - y2)
Trying to factor as a Difference of Squares:
4.2 Factoring: x2 - y2
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 : x2 is the square of x1
Check : y2 is the square of y1
Factorization is : (x + y) • (x - y)
Final result :
-2xy • (x + y) • (x - y)