Question

2r^2-16r+32...(2r+8)(r-4)??

Answer 1

can't be because if you multiply 8 by -4 you get a negative number

Answer 2

so it would be + sign then?

Answer 3

make your live easier by factoring out a 2 from the whole thing first, and write
\[2(r^2-8r+16)\]

Answer 4

so would it be (2r+8)(r+4)

Answer 5

then see if you can factor
\[r^2-8x+16\] as
\[(r-a)(r-b)\] both numbers have to be negative, because when you multiply them together you get +16 but they add up to - 8

Answer 6

no but close. they have to add up to -8, not +8

Answer 7

don't forget to factor out the 2 first, as i wrote, start with
\[2(r^2-8r+16)\]

Answer 8

2r^2-16r+32 ok thats the original equation

Answer 9

im lost on where i have to add up to -8

Answer 10

right, and your job is to factor. you see that each "coefficient" is even right? so factor out a 2 first

Answer 11

again make sure to start with
\[2(r^2-8r+16)\]

Answer 12

that is why i said you need two number whose product is +16 and whose sum is -8
in other words two numbers that when you multiply you get positive 16 but when you add you get minus 8

Answer 13

oh -2 and 4?

Answer 14

8 and 2?

Answer 15

not 8 and 2, because although 2 times 8 is 16, 2 + 8 is 10 and you need -8 not 10

Answer 16

4 and 4

Answer 17

getting warmer. 4 times 4 is 16 but 4 + 4 is 8
you want minus 8

Answer 18

so just make them both negative

Answer 19

in other words pick -4 and -4 because if you multiply them together you get
\[(-4)\times (-4)=+16\] and
\[-4+(-4)=-8\]

Answer 20

ugh i see now lol

Answer 21

and so your "final answer" is
\[2(r-4)(r-4)\]

Answer 22

more succinctly written as
\[2(r-4)^2\] impress your teacher by writing it with the exponent of 2.

Answer 23

ok i got 2(r-4)^2 lol i see where you answered it lol

Answer 24

good luck, and keep practicing!

Answer 25

thanks :)

Answer 26

yw

Answer 27

have a good night. I think im done lol ill be back tomorrow lol