Question

how do i factor 4k^3 +14k^2+ 14k + 4

Answer 1

any ideas? I didn't know if i should separate them into two problems

Answer 2

Hi :-)
make 14k common between middle terms and 4 between 4k^3 and 4

Answer 3

so would it become 4(k^3) + 14k(k+1)

Answer 4

you will come up with:
\[4(k^3+1)+14k(k+1)\]
then make use of this identity:\[a^3+b^3=(a+b)(a^2-ab+b^2)\]

Answer 5

I don't know how to use that identity

Answer 6

ok, I'll tell you :-) let \(a=k\) and \(b=1\) so you have:\[k^3+1=(k+1)(k^2-k+1)\]am i clear?

Answer 7

ok I am with you so far, what happens to the 4?

Answer 8

well, let's see what we have till here\[4(k+1)(k^2-k+1)+14k(k+1)\]now make 2(k+1) common, what will you get?

Answer 9

2(k+1)(k^2-k+1)(14k+4)

Answer 10

there is a mistake, give it a try again

Answer 11

(k+1)^2 (K^2-k+1) (14k+4)

Answer 12

Is it right this time?

Answer 13

sry, give me a minute, I'm coming

Answer 14

oh my bad

Answer 15

@hartnn plz help middi :-) i have to go

Answer 16

sure

Answer 17

\(4(k+1)(k^2-k+1)+14k(k+1)\)
factoring 2 and k+1

from the 1st term, what remains is 2 and k^2 -k+1
right ?

from the 2nd term, what remains is 7 and k

Answer 18

got this ?

\( 4(k+1)(k^2-k+1)+14k(k+1) \\ 2(k+1) [2 (k^2 -k+1) +7k] \)

Answer 19

you can simplify the terms inside [...] bracket

Answer 20

so would i start by applying 2 to what is inside the parenthesis

Answer 21

thats correct, do that

Answer 22

2k^2 +5k +1

Answer 23

just one error in that :

2k^2 -2k +2 +7k
= ...

Answer 24

whoops 2k^2 +5k + 2

Answer 25

yes
now thats easily factor-able
can you tell me 2 numbers with product = 2*2 and sum = 5 ?

Answer 26

(2k +1)(k+2)

Answer 27

correct!
now put everything together :)

Answer 28

2(k+1)(2k +1)(k+2)

Answer 29

2 (k+1) which we got previously
so,
\(\large 2 (2k+1)(k+1)(k+2)\)

Answer 30

correct! :)

Answer 31

thank you :)

Answer 32

welcome ^_^