Question

b(c-d)^2-a(d-c)+5c-5d​

Answer 1

Answer:

(d - c) • (-bc + bd + a + 3)

Step-by-step explanation:

Step 1 :

Equation at the end of step 1 :

((b•((c-d)2))+(a•(d-c)))+3•(c-d)

Step 2 :

Equation at the end of step 2 :

((b•((c-d)2))+a•(d-c))+3•(c-d)

Step 3 :

Equation at the end of step 3 :

(b•(c-d)2+a•(d-c))+3•(c-d)

Step 4 :

Pulling out like terms :

4.1 Pull out d-c

Note that c-d =(-1)• d-c

After pulling out, we are left with :

(d-c) • ( (-1) * (bc-bd-a) - 3 * (-1) ))

Trying to factor a multi variable polynomial :

4.2 Split -bc+bd+a+3

into two 2-term polynomials

+bd-bc and +a+3

This partition did not result in a factorization. We'll try another one:

-bc+bd and +a+3

This partition did not result in a factorization. We'll try another one:

-bc+a and +bd+3

This partition did not result in a factorization. We'll try another one:

-bc+3 and +a+bd

This partition did not result in a factorization. We'll try another one:

+3-bc and +a+bd

This partition did not result in a factorization. We'll try another one:

+a-bc and +bd+3

All three partitions failed. Tiger finds no factorization

Answer 2

Answer:

Step-by-step answer

given:-

b(c-d)2-a(d-c)+5c-5d

b (c-d)2-a(d-c)+5(c+d)

now here we can write -a(d-c) as a(c-d)

b(c-d)2+a(c-d)+5(c-d)

take (c-d) common

(c-d)[b(c-d)+a+5]

(c-d)[bc-bd +a+5]

hope helps explanation: