Verified answer

(7x+3)(3x+7)

x= -3/7, -7/3

(7x+3y)(3x+7y) is the answer... if you want proof, factor it out...

first=21x^2

outer=49xy

inner=9xy

outer+inner=58xy

last=21y^2

so, the polynomial is 21x^2+58xy+21y^2, just like you started with...

(7x+3y)(3x+7y)

The answer will be in the form

(ax + by) (cx + dy)

which is

(ac)x^2 + (bc+ad)xy + (bd)y^2

We need to find a, b, c and d such that

(ac) = 21

(bc+ad) =58 and

(bd) = 21

a and c must be either 3 and 7 or 21 and 1

likewise

b and d must be either 3 and 7 or 21 and 1

So, what combination gives us

(bc+ad) =58?

7x7 + 3x3

This makes a, b , c and d =

7, 3, 3, and 7 and

(ax + by) (cx + dy) =

(7x+3y)(3x+7y)

( 7x + 3) ( 3x + 7)

x= - 3/7 x= -7/3