Help me i need this today or will be kick out (y)+9)(2y-3)=0?

Question

it asked to solve the equation #1 (y)+9)(2y-3)=0 #2 m^2 +2m-15=0 #3 8r^2=24r #3 b^2-49=0 # 4 c(3c+16)=0 #5 z^4+8z^3 - 9z^2=0 #6 9y -7)(2y^2+7y-15)=0 plese help

Answer 1

This layout is really appalling and hard to read - please put each question on a separate line in future.

Whenever you have a product of two things being 0, either one or the other (or both, if that's possible) must be 0.

(y+9)(2y-3) = 0
=> (y+9) = 0 or (2y-3) = 0
=> y = -9 or y = 3/2.

The purpose of factorisation in this context is to split the quadratic up into a product of two factors so we can solve it easily.

m^2 + 2m - 15 = 0
=> (m+5) (m-3) = 0
=> m + 5 = 0 or m - 3 = 0
=> m = -5 or m = 3.

8r^2 = 24r
=> 8r^2 - 24r = 0
=> 8r(r-3) = 0
=> r = 0 or r = 3.

b^2 - 49 = 0 - a difference of perfect squares:
=> (b - 7) (b + 7) = 0
=> b = ±7.

c(3c+16) = 0
=> c = 0 or c = -16/3

z^4 + 8z^3 - 9z^2 = 0
=> z^2 (z^2 + 8z - 9) = 0
=> z^2 (z + 9) (z - 1) = 0
=> z = 0, -9 or 1.

(9y - 7) (2y^2 + 7y - 15) = 0
=> (9y - 7) (2y - 3) (y + 5) = 0
=> y = 9/7, 3/2 or -5.

Answer 2

Ok this is all null factor law stuff.

#1 (y+9)(2y-3) = 0

to make this easier to understand we will replace (y+9) with r and (2y - 3) with s

so:
r * s = 0

knowing this we know that either r or s has to be 0 because the only way to multiply 2 numbers together and get 0 is if one of them is 0.

so from this you need to realize that either (y+9) = 0 or
(2y-3)=0

so solve each equation seperatly

y + 9 = 0
y = -9

and

2y - 3 =0
2y = 3
y = 3/2

so y has 2 awnsers

y = 3/2, -9

all the rest of the questions are just more complecated versions of the same theory.

#2 m^2 +2m-15=0

to make this work it needs to be factorized

change the equation to m^2 + Am + B

to factorize we need to think 2 numbers that can be multiplied together to equal B that also add to equal A.

So with your equation we need numbers that can be multiplied to equal -15 that also add to equal 2

that would be 5 and -3

so you put it into the equation

(m + 5)(m-3) = 0

so using the null factor law above we know one of the sides must be zero so either

m + 5 = 0
m = -5

or

m - 3 = 0
m = 3

so m = 3, -5

the others all just need to be factorized and then have the null factor law applied to them. If you need any help factorizing just go to:

http://www.purplemath.com/modules/factquad.htm