Question

factor x^4+x^2+1(add and subtract x^2)

Answer 1

@phi

Answer 2

i need some help with these

Answer 3

I guess we should do what they suggest
x^4+x^2+1 + x^2 - x^2

or
x^4+2 x^2+1 - x^2

if we label x^2 as y , the first 3 terms give y^2 +2y + 1 which looks like a perfect square
can you continue?

Answer 4

if we subtract first we get-(x^2-x^4-x^2-1)

Answer 5

No, we do not do that.
start with
\[ x^4+x^2+1 \]
we can add zero to this expression and not change it \( x^2 - x^2 =0 \)
\[ x^4+x^2+1 + x^2 - x^2\]
or
\[ x^4+2x^2+1 - x^2 \]
now focus just on
\[ x^4+2x^2+1 \]
can you factor that ? (it is the same as y^2 + 2y + 1... which factors nicely)

Answer 6

x(x^3+2x)(x+1)

Answer 7

idk i'm messing up i'm bad at these

Answer 8

ok. can you factor
y^2 + 2y + 1
?

look at the last number, and list all pairs of factors
for example, if the last number was 6 we would list:
1,6
2,3

but in this case we have 1, and all we can list is
1,1

now look at the pairs on the list (here we have only 1 choice)
do they add up to the middle number (the 2 in 2y)
if so, they are our factors. we write
(y+1)(y+1)

Answer 9

yes

Answer 10

notice that if we rename x^2 to y we can do this

\[ x^4 + 2 x^2 + 1\\(x^2)^2 + 2 x^2 + 1 \\ y^2 + 2y +1\\(y+1)(y+1)\\(y+1)^2\]
now replace y with x^2 to get the answer
\[ (x^2+1)^2\]

Answer 11

now you have
\[ x^4+2 x^2+1 - x^2 \\ (x^2+1)^2 - x^2\]

that is a difference of squares
use this rule
\[ a^2 - b^2 = (a-b)(a+b) \]

in this problem, a = (x^2+1) and b=x

Answer 12

x^2+1^2-x^2=(x^2+1-x)(x^2+1+x)

Answer 13

ok but you should put parens in:
x^2+1^2-x^2
should be written
(x^2+1)^2-x^2

but so far, so good. people write the polynomial in standard form.. x^2 -x +1 (the x's are listed according to their exponent)
so the answer is
(x^2-x+1)(x^2+x+1)