Question

factorise:4y^4+y^4
Rd sharma question

Answer 1

Answer:

4y^4 + x^4

(2y^2)^2 - (ix^2)^2

(2y^2 + ix^2)(2y^2 - ix^2)

(ix^2 - (sqrt2*iy)^2)(sqrt2y - sqrtix )(sqrt2y + sqrtix )

where i^2 = -1

or

we know that

(x + y) = x^2 + y^2 + 2xy

(x - y)^2 = x^2 + y^2 - 2xy

like that

(x^2 + y^2)^2 = x^4 + y^4 + 2x^2y^2

x^4 + y^4 = (x^2 + y^2 )^2 - 2x^2y^2 --(1

like that

x^4 + y^4 = (x^2 - y^2)^2 + 2x^2y^2 --(2

on adding -(1 and -(2

2(x^4 + y^4) = (x^2 +y^2)^2 + (x^2 - y^2)^2

x^4 + y^4=( (x^2 +y^2)^2 + (x^2 - y^2)^2)/2

Answer 2

Let’s start from the left-hand side and multiply out the brackets. We find

(x2+2y2)2−4x2y2=x4+4x2y2+4y4−4x2y2=x4+4y4,

We know that

x4+4y4=(x2+2y2)2−4x2y2.

Notice that the right hand side is of the form a2−b2, where a=x2+2y2 and b=2xy.

We know that a2−b2=(a+b)(a−b), and so

(x2+2y2)2−(2xy)2=(x2+2y2+2xy)(x2+2y2−2xy).

We can think of the first bracket as

x2+2yx+2y2,

which is a quadratic in x with coefficients that depend on y. If this factorised into (x+ay)(x+by) for some a and b, then there would be one or two real solutions to x2+2yx+2y2=0 for each real value of y. So we can check the discriminant,

(2y)2−4(2y2)=4y2−8y2=−4y2.

But this is less than zero whenever y is non-zero. Thus x2+2y2+2xy=0 doesn’t have any real solutions if y≠0, and so we can’t factorise the first bracket any further.

For the second bracket, the discriminant is also

(−2y)2−4(2y2)=−4y2,

and so we can’t factorise the second bracket either. Therefore we’ve fully factorised x4+4y4.