Question

14(3y-5z)*2 - 21(5z-3y)*3 factorise the following by regrouping

Answer 1

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Answer:

14(3y-5z)^3+7(3y-5z)^2=7(3y - 5z)^2(6y - 10z + 1)14(3y−5z)3+7(3y−5z)2=7(3y−5z)2(6y−10z+1)

Step-by-step explanation:

Given : Expression 14(3y-5z)^3+7(3y-5z)^214(3y−5z)3+7(3y−5z)2

To find : Factories the expression ?

Solution :

To factor the expression we have to take common from terms,

14(3y-5z)^3+7(3y-5z)^2=7[2(3y - 5z)^3 + (3y - 5z)^2]14(3y−5z)3+7(3y−5z)2=7[2(3y−5z)3+(3y−5z)2]

14(3y-5z)^3+7(3y-5z)^2=7(3y - 5z)^2[2(3y - 5z) + 1]14(3y−5z)3+7(3y−5z)2=7(3y−5z)2[2(3y−5z)+1]

14(3y-5z)^3+7(3y-5z)^2=7(3y - 5z)^2[2(3y) - 2(5z) + 1]14(3y−5z)3+7(3y−5z)2=7(3y−5z)2[2(3y)−2(5z)+1]

14(3y-5z)^3+7(3y-5z)^2=7(3y - 5z)^2(6y - 10z + 1)14(3y−5z)3+7(3y−5z)2=7(3y−5z)2(6y−10z+1)

Therefore, 14(3y-5z)^3+7(3y-5z)^2=7(3y - 5z)^2(6y - 10z + 1)14(3y−5z)3+7(3y−5z)2=7(3y−5z)2(6y−10z+1)

Answer 2

Answer:

7(5z-3y)^2[2-15z+9y]

i hope it's help you