Factorization of x4 + x2y2 + y4 |
What you should know before… a2 – b2 = (a + b) (a – b) (1) a3 + b3 = (a + b) (a2 - ab + b2) (2) a3 – b3 = (a – b) (a2 + ab + b2) (3) a2 + 2ab + b2 = (a + b)2 (4) |
The Round-about Tour Let us begin with the factorization of x6 – y6 in two ways : (a) x6 – y6 = (x2)3 – (y2)3 = (x2 – y2)[(x2)2 + x2y2 +(y2)2] = (x + y)(x – y)(x4 + x2y2 + y4). (b) x6 – y6 = (x3)2 – (y3)2 = (x3 + y3)(x3 – y3) = (x + y)(x2 – xy + y2)(x - y)(x2 +xy +y2). Which of the above factorization is correct? Of course, (b) is the complete factorization, (a) is not. Comparing the results in (a) and (b), we can get: x4 + x2y2 + y4 = (x2 + xy + y2)(x2 –xy + y2) |
Further investigation x4 + x2y2 + y4 = (x4 + 2x2y2 + y4) - x2y2 = (x2 + y2)2 – (xy)2 = [(x2 + y2) + xy] [(x2 + y2) – xy] = (x2 + xy + y2)(x2 –xy + y2) |
Similar way There are some factorization which use the same technique, here is one example: x4 + 4 = (x4 + 4x2 + 4) – 4x2 = (x2 + 2)2 – (2x)2 = (x2 + 2 + 2x)(x2 + 2 – 2x) = (x2 + 2x + 2)(x2 – 2x + 2). |
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