3 years ago

(6z²+z-1)(9z-5) polynomial equation

1 answer:

8 0

Hi there, (6z²+z-1)(9z-5)=(6z²+z+-1)(9z+-5)=(6z²)(9z)+(6z²)(-5)+(z)(9z)+(z)(-5)+(-1)(9z)+(-1)(-5)=54z³-30z²+9z²-5z-9z+5=54z³-21z²-14z+5. Therefore, the answer is 54z³-21z²-14z+5.

You might be interested in

4 divided by 457. Please list the steps

exis [7]

You find out how many times 4 goes into 4 and get 1 then you find out how many times 4 goes into 5 and get 1 but you subtract 5 minus 4 and get 1 then bring down the 7 to get 17 then find out how many times 4 goes into 17 which is 4 times because 4 times 4 is 16 and you do 17 minus 16 and get 1 then add a decimal and bring down the 1 to get 10 then find out how many times 4 goes into 10 and get 2 subtract and get 2 bring down the zero turn it into 0 then you get 20 then find out how many times 4 goes into 20 and get 5 and 4*5=20 so 20-20 is 0 so your answer is 114.25

5 0

3 years ago

Read 2 more answers

Find (f×g)(x) when f(x)=x^2+8x+15 and g(x)=5 over x^2-9

Kipish [7]

Look at the picture......