# Further factorisation

• Dec 24th 2009, 04:49 PM
M.R
Further factorisation
I am stuck in this two question. I have gone as far as i can, and i provided the answer. So if someone can fill in the blank it would be great:

(a) $x^7 + x$
$= x (x^6 + 1)$
$= x ((x^2)^3 + 1^3)$
$= x (x^2 + 1) (x^4 - x^2 + 1)$
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$= x (x^2 + 1) (x^2 - \sqrt{3}x + 1) (x^2 + \sqrt{3}x + 1)$

(b) $x^{12} - y^{12}$
$= (x^4)^3 - (y^4)^3$
$= (x^4 - y^4) (x^8 + x^4 y^4 + y^8)$
$= (x^2 - y^2) (x^2 + y^2) (x^8 + x^4 y^4 + y^8)$
$= (x - y) (x + y) (x^2 + y^2) (x^8 + x^4 y^4 + y^8)$
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.
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= $(x - y) (x + y) (x^2 + y^2) (x^2 + xy + y^2) (x^2 - xy + y^2) (x^2 - \sqrt{3}xy + y^2) (x^2 + \sqrt{3}xy + y^2)$

Note: The questions are from Cambridge Mathematics 3 Unit (year 11). Exercise 1E, question 11
• Dec 24th 2009, 06:36 PM
Soroban
Hello, M.R!

Quote:

$(a)\;\;x^7 + x$
. . $= x (x^6 + 1)$
. . $= x ((x^2)^3 + 1^3)$
. . $= x (x^2 + 1) (x^4 - x^2 + 1)$

Answer: . $x (x^2 + 1) (x^2 - \sqrt{3}x + 1) (x^2 + \sqrt{3}x + 1)$

Normally, we could stop where you did.
. . Usually, "factoring" means integer coefficients.
But they pulled a cute trick and took it a step further.

That quartic polynomial, $x^4 - x^2 + 1$, can be factored
. . if we're desperate enough.

We have: . $x^4 - x^2 + 1$

Add and subtract $3x^2\!:\quad x^4 - x^2 \,{\color{red}+\, 3x^2} + 1 \,{\color{red}-\, 3x^2} \;=\;x^4 + 2x^2 + 1 - 3x^2$

So we have: . $(x^2+1)^2 - (\sqrt{3}x)^2$ . . . a difference of squares!

Factor: . $\left(x^2 + 1 - \sqrt{3}x\right)\left(x^2+1 + \sqrt{3}x\right) \;= \;\left(x^2 - \sqrt{3}x + 1\right)\left(x^2 + \sqrt{3}x + 1\right)$

• Dec 24th 2009, 07:29 PM
M.R
Thanks Soroban. And i have figured out part (b) also. Here it is:

(b) $x^{12} - y^{12}$
$= (x^4)^3 - (y^4)^3$
$= (x^4 - y^4) (x^8 + x^4 y^4 + y^8)$
$= (x^2 - y^2) (x^2 + y^2) (x^8 + x^4 y^4 + y^8)$
$= (x - y) (x + y) (x^2 + y^2) (x^8 + x^4 y^4 + y^8)$

Now as for $(x^8 + x^4 y^4 + y^8)$:
$=x^8+2x^4y^4+y^8 - x^4y^4$
$=(x^4+y^4)^2 - (x^2y^2)^2$
$=(x^4+y^4 - x^2y^2) (x^4+y^4 + x^2y^2)$

As for $=(x^4+y^4 + x^2y^2$:
$=x^4+2x^2y^2+y^4 - x^2y^2$
$=(x^2+y^2)^2 - (xy)^2$
$=(x^2-xy+y^2) (x^2+sy+y^2)$

As for $(x^4+y^4 - x^2y^2)$:
$=x^4+2x^2y^2+y^4 - 3x^2y^2$
$=(x^2+y^2)^2 - (\sqrt{3}xy)^2$
$=(x^2-\sqrt{3}xy+y^2) (x^2+\sqrt{3}xy+y^2)$

Putting it all together it =
$(x - y) (x + y) (x^2 + y^2) (x^2 + xy + y^2) (x^2 - xy + y^2) (x^2 - \sqrt{3}xy + y^2) (x^2 + \sqrt{3}xy + y^2)$