Factories: a^7b +ab^7.
only a brainlist genius can answer thi by full method
Answers
Answered by
12
a7b-ab7
Final result :
ab•(a+b)•(a2-ab+b2)•(a-b)•(a2+ab+b2)
Step by step solution :
Step 1 :
Step 2 :
Pulling out like terms :
2.1 Pull out like factors :
a7b - ab7 = ab • (a6 - b6)
Trying to factor as a Difference of Squares :
2.2 Factoring: a6 - b6
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 : a6 is the square of a3
Check : b6 is the square of b3
Factorization is : (a3 + b3) • (a3 - b3)
Trying to factor as a Sum of Cubes :
2.3 Factoring: a3 + b3
Theory : A sum of two perfect cubes, a3 + b3can 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 : a3 is the cube of a1
Check : b3 is the cube of b1
Factorization is :
(a + b) • (a2 - ab + b2)
Trying to factor a multi variable polynomial :
2.4 Factoring a2 - ab + b2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Trying to factor as a Difference of Cubes:
2.5 Factoring: a3 - b3
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 : a3 is the cube of a1
Check : b3 is the cube of b1
Factorization is :
(a - b) • (a2 + ab + b2)
Trying to factor a multi variable polynomial :
2.6 Factoring a2 + ab + b2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Final result :
ab•(a+b)•(a2-ab+b2)•(a-b)•(a2+ab+b2)
Final result :
ab•(a+b)•(a2-ab+b2)•(a-b)•(a2+ab+b2)
Step by step solution :
Step 1 :
Step 2 :
Pulling out like terms :
2.1 Pull out like factors :
a7b - ab7 = ab • (a6 - b6)
Trying to factor as a Difference of Squares :
2.2 Factoring: a6 - b6
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 : a6 is the square of a3
Check : b6 is the square of b3
Factorization is : (a3 + b3) • (a3 - b3)
Trying to factor as a Sum of Cubes :
2.3 Factoring: a3 + b3
Theory : A sum of two perfect cubes, a3 + b3can 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 : a3 is the cube of a1
Check : b3 is the cube of b1
Factorization is :
(a + b) • (a2 - ab + b2)
Trying to factor a multi variable polynomial :
2.4 Factoring a2 - ab + b2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Trying to factor as a Difference of Cubes:
2.5 Factoring: a3 - b3
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 : a3 is the cube of a1
Check : b3 is the cube of b1
Factorization is :
(a - b) • (a2 + ab + b2)
Trying to factor a multi variable polynomial :
2.6 Factoring a2 + ab + b2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Final result :
ab•(a+b)•(a2-ab+b2)•(a-b)•(a2+ab+b2)
yoyoluckysingh:
ok
Similar questions