algebra help please?

Question

I need to use factoring to solve the equation 8 z ^2 = 56 z - 48, I think the answer possibly is z= - 6, and z = -1 but I am not too sure can anyone help?

Answer 1

first set the equation =0
8z^2-56z+48=0
now factor out 8
8(z^2-7z+6)=0
you can forget about the 8 and just set z^2-7z+6=0 and factor:
(z-1)(z-6)=
z-1=0
z=1

z-6=0
z=6

Answer 2

Rewrite this as 8z^2 - 56z + 48 = 0
You can factor out an 8 which will make things easier:
8(x^2 - 7z + 6) = 0

Now factor the rest. We have a - before the 6 and a minus before the 7z, so we need factors in the form of (z - ?)(z - ?). We just need two numbers whose product is 6 and sum is 7. The obvious ones are 1 and 6. This gives us:
8(x - 6)(x - 1) = 0

So x = 6, 1

Answer 3

First, move all the terms to one side:
8z^2-56z+48=0

Next, divide both side by 8 and you get:
z^2-7z+6=0

The factors of 6 are 1 & 6 or 2 & 3. factoring yields:
(z-1)(z-6)=0 Therefore z = 1 or 6

Answer 4

8z^2 - 56z + 48 = 0
dividing by 8 yields
z^2 - 7z + 6 = 0
Factoring:
(z-?)(z-?) = 0 (rule states that the second sign is positive when both factors have a minus sign; the first sign is negative, thus both signs must be negative)
(z-6)(z-1) = 0
(The only factorization of 6 that adds to 7 is 6 and 1).
Thus z = 1 or 6

Answer 5

okay.. um... first of all you want to move all the terms to one side. so you subtract the 56z and add the 48 to the left to make 0 on the right. Okay.. then you will have 8z^2-56z+48=0 right?... Okay.. the easiest thing to do is to divide all the terms, even 0, by 8, the least common multiple. so you will get z^2-7z+6=0. now the problem doesnt look as scary as it did before right? then you factor... you get (z-6) (z-1)=0. therefore.. z=6 and 1. if you have any questions.. feel free to email me at [e-mail]

Answer 6

8 z² - 56z + 48 = 0

Divide thru by 8

z² - 7z + 6 = 0

factor

(z - 1)(z - 6) = 0

z = 1, z = 6
.

Answer 7

first, transpose 56z-48 to the other side:
8z^2-56z+48=0

then factor:
(4z-4)(2z-12)=0

4z-4=0
4z=4
z=1

2z-12=0
2z=12
z=6