Question

Factorise z(5z^2-80)÷5z(z+4)

Answer 1

Answer:

Mark this as brainliest please

Step-by-step explanation:

z(5z²-80)/5z(z+4)

Z will cancel out from numerator and denominator

So

5z²-80/5(z+4)

Take 5 common from numerator and cancel it with denominator

Z²-16/z+4

Factorise numerator using the identity a²-b²=(a+b)(a-b)

Z-4 is your answer

Answer 2

z(5z2-80)/5z(z+4) 

Final result :

z2 • (z + 4)2 • (z - 4)

Step by step solution :

Step  1  :

Equation at the end of step  1  :

(5z2 - 80) ((z • ——————————) • z) • (z + 4) 5

Step  2  :

5z2 - 80 Simplify ———————— 5

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   5z2 - 80  =   5 • (z2 - 16) 

Trying to factor as a Difference of Squares :

 3.2      Factoring:  z2 - 16 

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 : 16 is the square of 4
Check :  z2  is the square of  z1 

Factorization is :       (z + 4)  •  (z - 4) 

Equation at the end of step  3  :

((z • (z + 4) • (z - 4)) • z) • (z + 4)

Step  4  :

Equation at the end of step  4  :

(z • (z + 4) • (z - 4) • z) • (z + 4)

Step  5  :

Multiplying exponential expressions :

 5.1    z1 multiplied by z1 = z(1 + 1) = z2

Equation at the end of step  5  :

z2 • (z + 4) • (z - 4) • (z + 4)

Step  6  :

Multiplying Exponential Expressions :

 6.1    Multiply  (z+4)  by  (z+4) 

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (z+4)  and the exponents are :
          1 , as  (z+4)  is the same number as  (z+4)1 
 and   1 , as  (z+4)  is the same number as  (z+4)1 
The product is therefore,  (z+4)(1+1) = (z+4)2 

Final result :

z2 • (z + 4)2 • (z - 4)

make it brainliest please