challenge question -- Factor the polynomial completely
- Thread starter lookagain
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Method 1:
Factor first by grouping:
\(\displaystyle (x^5+x^4)+(x^3+x^2)+(x+1)=x^4(x+1)+x^2(x+1)+(x+1)=(x+1)(x^4+x^2+1)\)
Now, for the quartic factor assume it may be factored as follows:
\(\displaystyle x^4+x^2+1=(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1\)
Equating coefficients, we find:
\(\displaystyle a+b=0\)
\(\displaystyle ab+2=1\)
and so one solution is \(\displaystyle (a,b)=(1,-1)\) and we have:
\(\displaystyle x^4+x^2+1=(x^2+x+1)(x^2-x+1)\) which means:
\(\displaystyle x^5+x^4+x^3+x^2+x+1=(x+1)(x^2+x+1)(x^2-x+1)\)
Method 2:
Let:
\(\displaystyle S=x^5+x^4+x^3+x^2+x+1\) and so:
\(\displaystyle Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1\) hence:
\(\displaystyle S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)\) thus:
\(\displaystyle S=(x+1)(x^2-x+1)(x^2+x+1)\)
Factor first by grouping:
\(\displaystyle (x^5+x^4)+(x^3+x^2)+(x+1)=x^4(x+1)+x^2(x+1)+(x+1)=(x+1)(x^4+x^2+1)\)
Now, for the quartic factor assume it may be factored as follows:
\(\displaystyle x^4+x^2+1=(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1\)
Equating coefficients, we find:
\(\displaystyle a+b=0\)
\(\displaystyle ab+2=1\)
and so one solution is \(\displaystyle (a,b)=(1,-1)\) and we have:
\(\displaystyle x^4+x^2+1=(x^2+x+1)(x^2-x+1)\) which means:
\(\displaystyle x^5+x^4+x^3+x^2+x+1=(x+1)(x^2+x+1)(x^2-x+1)\)
Method 2:
Let:
\(\displaystyle S=x^5+x^4+x^3+x^2+x+1\) and so:
\(\displaystyle Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1\) hence:
\(\displaystyle S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)\) thus:
\(\displaystyle S=(x+1)(x^2-x+1)(x^2+x+1)\)
D
Deleted member 4993
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(x+1)(x4+x2+1) → (x+1)(x4+2x2+1 - x2) → (x+1)(x2+1+x)(x2+1-x)
or
(x3+1)(x2+x+1) →(x+1)(x2-x+1)(x2+x+1)
Here is another:
(x^5 + 1) + (x^4 + x^3 + x^2 + x) =
(x + 1)(x^4 - x^3 + x^2 - x + 1) + (x^4 + x^3) + (x^2 + x) =
(x + 1)(x^4 - x^3 + x^2 - x + 1) + x^3(x + 1) + x(x + 1) =
(x + 1)(x^4 - x^3 + x^2 - x + 1 + x^3 + x) =
(x + 1)(x^4 + x^2 + 1) =
(You can continue simplifying this with one of the above steps in
any of the appropriate above posts.)
(x^5 + 1) + (x^4 + x^3 + x^2 + x) =
(x + 1)(x^4 - x^3 + x^2 - x + 1) + (x^4 + x^3) + (x^2 + x) =
(x + 1)(x^4 - x^3 + x^2 - x + 1) + x^3(x + 1) + x(x + 1) =
(x + 1)(x^4 - x^3 + x^2 - x + 1 + x^3 + x) =
(x + 1)(x^4 + x^2 + 1) =
(You can continue simplifying this with one of the above steps in
any of the appropriate above posts.)
Last edited:
Here's another way.
Since \(\displaystyle x^5+x^4+x^3+x^2+x+1 = \dfrac{x^6-1}{x-1}\) the roots of this polynomial are exactly the set \(\displaystyle \{z\in \mathbb{C}-\{1\}\,\,;\,\, z^6=1\}\), i.e. the roots of unity, ignoring the positive real root. They are \(\displaystyle e^{\pm i\pi/3},e^{\pm i2\pi/3}, -1 \).
We want a real factorization obviously, and we can see that the conjugate pair to each root is present (as it should be). Pairing them off we get the (minimal) polynomials for each:
\(\displaystyle (x-e^{i\pi/3})(x-e^{- i\pi/3}) = x^2-x+1\)
\(\displaystyle (x-e^{i2\pi/3})(x-e^{- i2\pi/3}) = x^2+x+1\)
\(\displaystyle x-(-1) = x+1\)
There is an abstract algebra/number theoretic variation of the above that can be performed for the general case too.
Since \(\displaystyle x^5+x^4+x^3+x^2+x+1 = \dfrac{x^6-1}{x-1}\) the roots of this polynomial are exactly the set \(\displaystyle \{z\in \mathbb{C}-\{1\}\,\,;\,\, z^6=1\}\), i.e. the roots of unity, ignoring the positive real root. They are \(\displaystyle e^{\pm i\pi/3},e^{\pm i2\pi/3}, -1 \).
We want a real factorization obviously, and we can see that the conjugate pair to each root is present (as it should be). Pairing them off we get the (minimal) polynomials for each:
\(\displaystyle (x-e^{i\pi/3})(x-e^{- i\pi/3}) = x^2-x+1\)
\(\displaystyle (x-e^{i2\pi/3})(x-e^{- i2\pi/3}) = x^2+x+1\)
\(\displaystyle x-(-1) = x+1\)
There is an abstract algebra/number theoretic variation of the above that can be performed for the general case too.
Hello, lookagain
This is a variation of daon's solution.
\(\displaystyle P(x) \;=\;\dfrac{x^6 - 1}{x-1} \;=\;\dfrac{\overbrace{(x^3)^2 - (1^2)}^{\text{diff. of squares}}}{x-1} \)
. . . . .\(\displaystyle =\;\dfrac{\overbrace{(x^3-1)}^{\text{diff.of cubes}}\cdot\overbrace{(x^3+1)}^{\text{sum of cubes}}}{x-1}\)
. . . . .\(\displaystyle =\; \dfrac{(\color{red}{\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\color{red}{\rlap{/////}}x-1}\)
. . . . .\(\displaystyle =\; (x+1)(x^2+x+1)(x^2-x+1)\)
This is a variation of daon's solution.
\(\displaystyle P(x) \;=\;\dfrac{x^6 - 1}{x-1} \;=\;\dfrac{\overbrace{(x^3)^2 - (1^2)}^{\text{diff. of squares}}}{x-1} \)
. . . . .\(\displaystyle =\;\dfrac{\overbrace{(x^3-1)}^{\text{diff.of cubes}}\cdot\overbrace{(x^3+1)}^{\text{sum of cubes}}}{x-1}\)
. . . . .\(\displaystyle =\; \dfrac{(\color{red}{\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\color{red}{\rlap{/////}}x-1}\)
. . . . .\(\displaystyle =\; (x+1)(x^2+x+1)(x^2-x+1)\)
MarkFL said:
These two (MarkFL's and soroban's versions) look essentially the same to me.
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Others:
(x^5 + x^2) + (x^4 + x) + (x^3 + 1) =
x^2(x^3 + 1) + x(x^3 + 1) + 1(x^3 + 1) =
(x^3 + 1)(x^2 + x + 1) =
(x + 1)(x^2 - x + 1)(x^2 + x + 1)
. . . . . . . . . . . . . . . . . . . . .
(x^5 + x^3 + x) + (x^4 + x^2 + 1) =
x(x^4 + x^2 + 1) + 1(x^4 + x^2 + 1) =
(x^4 + x^2 + 1)(x + 1) =
(x^2 - x + 1)(x^2 + x + 1)(x + 1)