Question

Polynomial help :p

Answer 1

\[n ^{4}+6n ^{3}+11n ^{2}+6n+1=(an ^{2}+bn+c)(pn ^{2}+qn+r)\]

Answer 2

\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
you're needing to factor this, right?
\(\color{#0cbb34}{\text{End of Quote}}\)
hehe, yessir

Answer 3

Also I have to show my work :P

Answer 4

uh.. try breaking it down..

Answer 5

\(\color{#0cbb34}{\text{Originally Posted by}}\) @animeobsessionn
uh.. try breaking it down..
\(\color{#0cbb34}{\text{End of Quote}}\)
Oh um thx , I thought I had to build up smh .-.

Answer 6

For the values of a and c you just need to look at the highest coefficient (This case is \(n^4\))

\(an^2\times pn^2=n^4\)

Do you know what the values of a and c would be here?

Answer 7

No-

Answer 8

Wait is it 1?

Answer 9

\(an^2\times pn^2=1n^4\)

Just looking at the coefficients...

\(a\color{red}{n^2}\times p\color{red}{n^2}=1\color{red}{n^4}\)

\(a\times p=1\)
What times what is 1?

Answer 10

Yes its 1

Answer 11

1*1=1 ._.

Answer 12

Now we move on to b and q

For this we are looking at the ones with the power of 3. Which is only \(6n^3\)

You want \(bn\times n^2=3n^3\) \(==> b\times 1=3\)
and for q: \(qn\times n^2=3n^3\) \(==> q\times 1=3\)
What times 1 is 3?

Answer 13

1x3=3 ._.

Answer 14

Yes so b and q would be 3


Now for the last two, c and r

Because this is the constant, you just have to look at the last value \(n^4+6n^3+11n^2+6n\color{green}{+1}\)

\(c\times r=1\)?

Answer 15

Errrrrrmmmmmm yes?

Answer 16

What times what is 1? xD

Answer 17

Errrrrmmmm no? xd rip me uh 1*1 is 1?

Answer 18

Yes, so c and r is 1

\((n^2+3n+1)(n^2+3n+1)\)

Answer 19

If you want to check it

\((n^2+3n+1)(n^2+3n+1)\)

\(n^4+3n^3+3n^3+n^2+n^2+9n^2+3n+3n+1\)

\(n^4+6n^3+11n^2+6n+1\)

Answer 20

\(\color{#0cbb34}{\text{Originally Posted by}}\) @dude
If you want to check it

\((n^2+3n+1)(n^2+3n+1)\)

\(n^4+3n^3+3n^3+n^2+n^2+9n^2+3n+3n+1\)

\(n^4+6n^3+11n^2+6n+1\)
\(\color{#0cbb34}{\text{End of Quote}}\)
Umm ok so they are equal right?

Answer 21

Yes

Answer 22

LMAO i understood absulutely none of that,expect for the part where r and c equals now
O-o

Answer 23

\(\color{#0cbb34}{\text{Originally Posted by}}\) @IhelpUuHELPme
LMAO i understood absulutely none of that,expect for the part where r and c equals now
O-o
\(\color{#0cbb34}{\text{End of Quote}}\)
Lucky it isnt your hw

Answer 24

\(\color{#0cbb34}{\text{Originally Posted by}}\) @boredfr
\(\color{#0cbb34}{\text{Originally Posted by}}\) @IhelpUuHELPme
LMAO i understood absulutely none of that,expect for the part where r and c equals now
O-o
\(\color{#0cbb34}{\text{End of Quote}}\)
Lucky it isnt your hw
\(\color{#0cbb34}{\text{End of Quote}}\)
no duh im lucky XD