# Answer to Question #11093 in Algebra for nistala ks prasad

Question #11093

factorize

m^8-11m^4n^4-80n^8

m^8-11m^4n^4-80n^8

Expert's answer

A = m^8-11m^4n^4-80n^8 = n^8 ( (m/n)^8 - 11 (m/n)^4 - 80 )

Replace: t=(m/n)^4.

Then

A = n^8 ( t^2 - 11t - 80 )

Let us solve

It is easy to verify that the equation:

t^2 - 11t - 80 = 0

has the following two roots:

t1 = 16

t2 = -5.

Hence

t^2 - 11t - 80 = (t-16)(t+5).

Thus

A = n^8 (t-16)(t+5)

= n^8 * ( (m/n)^4 - 16 ) * ( (m/n)^4 + 5 )

= ( m^4 - 16 n^4 ) * ( m^4 + 5 n^4 ) =

= ( (m^2)^2 - (4n^2)^2 ) * ( m^4 + 5 n^4 ) =

= (m^2-4n^2) * (m^2 + 4n^2) * ( m^4 + 5 n^4 ) =

= ( m^2 - (2n)^2 ) * (m^2 + 4n^2) * ( m^4 + 5 n^4 ) =

= (m-2n) * (m+2n) * (m^2 + 4n^2) * ( m^4 + 5 n^4 ).

Thus

m^8 - 11m^4n^4 - 80n^8 = (m-2n) * (m+2n) * (m^2 + 4n^2) * ( m^4 + 5 n^4 ).

Replace: t=(m/n)^4.

Then

A = n^8 ( t^2 - 11t - 80 )

Let us solve

It is easy to verify that the equation:

t^2 - 11t - 80 = 0

has the following two roots:

t1 = 16

t2 = -5.

Hence

t^2 - 11t - 80 = (t-16)(t+5).

Thus

A = n^8 (t-16)(t+5)

= n^8 * ( (m/n)^4 - 16 ) * ( (m/n)^4 + 5 )

= ( m^4 - 16 n^4 ) * ( m^4 + 5 n^4 ) =

= ( (m^2)^2 - (4n^2)^2 ) * ( m^4 + 5 n^4 ) =

= (m^2-4n^2) * (m^2 + 4n^2) * ( m^4 + 5 n^4 ) =

= ( m^2 - (2n)^2 ) * (m^2 + 4n^2) * ( m^4 + 5 n^4 ) =

= (m-2n) * (m+2n) * (m^2 + 4n^2) * ( m^4 + 5 n^4 ).

Thus

m^8 - 11m^4n^4 - 80n^8 = (m-2n) * (m+2n) * (m^2 + 4n^2) * ( m^4 + 5 n^4 ).

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