algebra homework! not getting it at all?

Question

problems as follows
(3z+5)(2z-7)
(2-y)(4-y)

Answer 1

Note: the symbol * means multiplication

1/(3z+5)(2z-7)
============
(3z+5)(2z-7) is a short notation for (3*z+5)*(2*z-7)

This is exactly equivalent to (3*z)*(2*z-7) +(5)*(2*z-7)
This is exactly equivalent to 3*z*2*z +3*z*(-7) +5*2*z +5*(-7)

now lets reorganize all that:
3*2*z*z -3*7*z +5*2*z -5*7
so in the end we get
6*z*z - 21*z +10*z -35
we then sum the z
6*z*z -11*z-35

or in a more compact notation: 6z^2-11z-35

2/ (2-y)(4-y)
==========
This means (2-y)*(4-y)
you do like we did in 1/ and you get y^2-6y+8

Answer 2

I'm assuming you want to expand both of these. Many people use the FOIL method, although I didn't as I tended to make too many simple mistakes and stuff up my answers. I write them out in full to work it out. You will have to find out which method works best for you.

You have to use these kinds of methods as you have to times everything in the first bracket by everything in the second one!

=(3z+5)(2z-7)
=3z(2z-7)+5(2z-7)
=6z^2-21z+10z-35

=6z^2-11z-35

Same for the other one! Hope this helps!

Answer 3

what do you want ?
i do product

(3z+5) (2z-7) = 6z^2-29z -35


(2-y) (4-y) = y^2 -6y +8


if you want the roots first z = -5/3 z= 7/2 first expression

y= 2 and y=4 second expression

Answer 4

You need to FOIL the binomials:

First - (3z * 2z)
Outer - (3z * -7)
Inner - (5 * 2z)
Last - (5 * -7)

Add these products and you get your answer. Do the same for the other one.

Answer 5

6z^2-21z+10z-35
answer: 6z^2-11z-35

8-2y-4y+y^2
answer: 8-6y+y^2

Answer 6

Just expand them