Factoring + Homework?

Question

One more time?

How do you do this?

Please and thank you.

r[2 the second power] - 24r + 44

0 = r [to the second power] - 11r - 60

(x-5)(3-x) = 0

Answer 1

r² - 24r + 44

(r-22)(r-2)

x= 22, x= 2

r² - 11r - 60

(r-15)(r+4)

x= 15, x= -4

(x-5)(3-x)

(x-5)(-x+3)

-(x-3)(x-5)

x= 5, x= 3

Answer 2

r^2-24r+44= ( r - 22 )(r  -2 )   In order to do this you have to think what two numbers mulitply to get 44.  That would be the pairs, 1,44  -1,-44

                                                                                    2,22  -2,-22

                                                                                    4,11  -4,-11.

Ask which pairs adds to give you -24.  There will always be just one right answer.

2)  0=r^2-11r-60

     0=( r  -15 )(  r   + 4  )    

   Now, that means either r-15=0 or r+4=0

                                      r=15 or r= -4

In order to factor that one, think what numbers add to give  -60

  -1, 60 or 1, -60

  -2, 30 or  2, -30

-3, 20 or 3, -20

-4. 15 or 4, -15

-5,12  or 5, -12

-6,10 or 6, -10     The only pairs that adds to give you -11 is -15,4

3) Either x-5 =0 or 3-x=0 because if two numbers multiply to give you zero, one of them has to be zero, so solve each little equation and get xx=5 or x=3

Hope that helped a little.

Answer 3

Okay:

(r+2)(r-22) = 0.  Set each term = 0 and solve for each, r = -2 and 22

0 = (r+4)(r-15) and same as above, r = -4 and 15

set each = 0, x = 5 or -3  

Answer 4

r^2 - 24r + 44 = (r-2)(r-22)

r^2 - 11r - 60 = 0

(r-15)(r+4) = 0

r = 15 or -4

(x-5)(3-x)=0

x= 5 or 3