Question

how do you simplify this without multiplying them all together? any help is appreciated

(2x)^5 (5) (x^2+x+1)^4 (2x+1) + (x^2+x+1)^5 (5) (2x)^4 (2)

Answer 1

convert into whole squares

Answer 2

how do you do that?

Answer 3

sry try taking common like (2x)^4 out of the whole exp

Answer 4

it looks maybe like binomial stuff

Answer 5

Factor out the greatest common factor:
Both have a 5 in common
Both have (2x)^4 in common
Both have (x^2+x+1)^4 in common

5(2x)^4(x^2+x+1)^4[2x(2x+1) + 2(x^2 +x+1)]

Answer 6

then simplify, add like terms...

Answer 7

im prolly wrong about that tho :)

Answer 8

okay!! so then would you distribute the 2x and (2x+1) and then the 2 with the (x^2+x+1)??

Answer 9

yes

Answer 10

so after you do that would you just take out the gcf out of that and thats it?

Answer 11

yep
if you expand the (2x)^4 as well to 16x^4
you should get
\[160x^{4}(x^{2}+x+1)^{4}(3x^{2}+2x+1)\]

Answer 12

okay!! thank you so much! could i ask you one more..its a division problem and i want to make sure im doing it right

Answer 13

ok sure

Answer 14

\[(2)(3-x)(2x)-x ^{2}(2)(-1)\div [2(3-x)]^{2}\]

and obviously the denominator is supposed to be on the bottom

Answer 15

hello?

Answer 16

well first you can simplify the top by multiplying 2 and 2x, 2 and -1
\[= \frac{4x(3-x) + 2x^{2}}{4(3-x)^{2}}\]

Answer 17

You can factor out a 2 on top
resulting in
\[\frac{2x(3-x) + x^{2}}{2(3-x)^{2}}\]

Answer 18

can the (3-x) on top become a 1 and then the (3-x)^2 become just (3-x)?

Answer 19

the top can be then simplified to x(6-x)
if you distribute and add like terms
6x-2x^2 +x^2 = 6x - x^2 = x(6-x)

No they dont cancel because there is addition in the numerator

Answer 20

i think thats as far as you can go

Answer 21

how did the 2x^2 just become x^2?

Answer 22

oh i added -2x^2 + x^2 = -x^2
-2 +1 = -1

Answer 23

oh ok i didnt see that! and on this other expression
\[(x+7)^{3}(4)(3x+1)^{3}(3) + 3(x+7)^{2}(3x+1)^{4}\]
can you factor out that 3 or no because the 3 on the right is eventually distributed?

Answer 24

Yes it factors out
it doesn't matter where the 3 is as long as its being multiplied by the rest of the term

Answer 25

ok!!

Answer 26

\[3(x+7)^{2}(3x+1)^{3}(4(x+7)+(3x+1))\]
\[=3(x+7)^{2}(3x+1)^{3}(7x+29)\]

Answer 27

thats what i got! and for (x-5)-(x+7) / (x-5)^2 do you distribute the bottom?