Question

x4+x2y2+y4 factorise

Answer 1

Answer:

Step-by-step explanation:

This can  be factored by completing the square. 

Recall that perfect square trinomial  x²+2xy+y² = (x+y)².

x⁴ + x²y² +y⁴                      first term and 3rd term  are perfect squares

= (x²)² +  x²y² + (y²)²          the middle term should to be 2x²y²  to make                                                              it a perfect square, so add x²y² - x²y² 

= (x²)² +  x²y² + (y²)²  + x²y² - x²y²       combine 2nd and 4th terms

= (x²)² +  2 x²y² + (y²)²  - x²y²      first 3 terms form a perfect square trionomial

= (x² + y²)²- (xy)²                   it is a difference of two squares, now factor

=(x² + y² + xy)(x² + y² - xy)

Hope it helps you my dear friend..

Answer 2

Hey mate,

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], by (3)

= (x + y)(x – y)(x4 + x2y2 + y4), by (1)

(b) x6 – y6 = (x3)2 – (y3)2 = (x3 + y3)(x3 – y3), by (1)

= (x + y)(x2 – xy + y2)(x - y)(x2 +xy +y2), by (2) and (3)

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 (note that one term is added and subtracted)

= (x2 + y2)2 – (xy)2, , by (4)

= [(x2 + y2) + xy] [(x2 + y2) – xy] , by (1)

= (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).

Hope it will help you.

It's M.S.V.