Can someone please answer this? I don't understand
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Final result : (4x + 5y) • (16x2 - 20xy + 25y2)
Step by step solution :Step 1 :Equation at the end of step 1 : (64 • (x3)) + 53y3 Step 2 :Equation at the end of step 2 : 26x3 + 53y3 Step 3 :Trying to factor as a Sum of Cubes :
3.1 Factoring: 64x3+125y3
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 : 64 is the cube of 4
Check : 125 is the cube of 5
Check : x3 is the cube of x1
Check : y3 is the cube of y1
Factorization is :
(4x + 5y) • (16x2 - 20xy + 25y2)
Trying to factor a multi variable polynomial :
3.2 Factoring 16x2 - 20xy + 25y2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Final result : (4x + 5y) • (16x2 - 20xy + 25y2)
Step by step solution :Step 1 :Equation at the end of step 1 : (64 • (x3)) + 53y3 Step 2 :Equation at the end of step 2 : 26x3 + 53y3 Step 3 :Trying to factor as a Sum of Cubes :
3.1 Factoring: 64x3+125y3
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 : 64 is the cube of 4
Check : 125 is the cube of 5
Check : x3 is the cube of x1
Check : y3 is the cube of y1
Factorization is :
(4x + 5y) • (16x2 - 20xy + 25y2)
Trying to factor a multi variable polynomial :
3.2 Factoring 16x2 - 20xy + 25y2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Final result : (4x + 5y) • (16x2 - 20xy + 25y2)
niya25:
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answer is (4x)power3 + (√5y)power3 =( 4x+√5y){(4x)power2 + (√5y)power2 + 4√5xy}
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