SOLUTION: I need some help with some homework Please!!!! The directions say: Solve the following polynomial equations using the zero factor property 3x^2 + 5x-2=0 2y^2-11y

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I need some help with some homework Please!!!! The directions say: Solve the following polynomial equations using the zero factor property 3x^2 + 5x-2=0 2y^2-11y      Log On


   



Question 146464: I need some help with some homework Please!!!!
The directions say: Solve the following polynomial equations using the zero factor property


3x^2 + 5x-2=0


2y^2-11y-40=0 thanks so much!!!!

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
I need some help with some homework Please!!!!
The directions say: Solve the following polynomial equations using the zero factor property


3x%5E2+%2B+5x-2+=+0 

Multiply the first number, 3, on the left by the last
number on the left, 2. (Take these as positive).

You get 6.  So we write down all the ways to get
two integers that have product 6.  These are

 products   
1 x 6 = 6    
2 x 3 = 6    

Now look at the sign of the last number in
3x%5E2+%2B+5x-2 = 0, -2.  It is negative, so out beside 
the multiplication, subtract the smaller of the 
factors from the larger factor: 

[Note: if that last sign had been positive, we would add,
not subtract.]

 products   differences
1 x 6 = 6    6 - 1 = 5
2 x 3 = 6    3 - 2 = 1 

3x%5E2+%2B+5x-2 = 0

Since the middle number 5 appears as a difference
between 6 and 1, we use 6 and 1 to rewrite the
middle term "5x".

We rewrite 5x as 6x+-+1x

So 3x%5E2+%2B+5x-2 = 0

becomes

3x%5E2%2B6x-1x-2 = 0

Now factor 3x out of the first two terms on the left:

3x%28x%2B2%29-1x-2 = 0

Now factor -1 out of the last two terms on the left:

3x%28x%2B2%29-1%28x%2B2%29 = 0

Now factor out the binomial factor %28x%2B2%29

%28x%2B2%29%283x-1%29 = 0

Set x%2B2=0 which gives solution x=-2

Set 3x-1=0 which gives solution x=1%2F3

--------------------------------------------
 
2y%5E2+-+11y-40=0  

Multiply the first number, 2, on the left by the last 
number on the left, 40.
(Take these as positive).

You get 80.  So we write down all the ways to get
two integers that have product 80.  These are

 products   
1 x 80 = 80    
2 x 40 = 80    
4 x 20 = 80
5 x 16 = 80
8 x 10 = 80

Now look at the sign of the last number in
2y%5E2+-+11y-40 = 0, -2.  It is negative, so out beside 
the multiplication, subtract the smaller of the 
factors from the larger factor: 

[Note: if that last sign had been positive, we would add,
not subtract.]

 products      differences
1 x 80 = 80    80 - 1 = 79    
2 x 40 = 80    40 - 2 = 38
4 x 20 = 80    20 - 4 = 16
5 x 16 = 80    16 - 5 = 11
8 x 10 = 80    10 - 8 = 2

2y%5E2+-+11y-40 = 0

Since the middle number 11 appears as a difference
between 16 and 5, we use 16 and 5 to rewrite the
middle term "-11y".

We rewrite -11y as -16y+%2B+5y

So 2y%5E2-11y-40 = 0

becomes

2y%5E2-16y%2B5y-40 = 0

Now factor 2y out of the first two terms on the left:

2y%28y-8%29%2B5y-40 = 0

Now factor +5 out of the last two terms on the left:

2y%28y-8%29%2B5%28y-8%29 = 0

Now factor out the binomial factor %28y-8%29

%28y-8%29%282y%2B5%29 = 0

Set y-8=0 which gives solution y=8

Set 2y%2B5=0 which gives solution y=-5%2F2

Edwin