Formula used:

The factors of a polynomial can be find by taking a common factor and this method is called factor by grouping,

\(ab+ac+bd+cd=a(b+c)+d(b+c)\)

\(=(a+d)(b+c)\)

Or,

\(ab - ac + bd-cd = a(b-c)+d(b-c)\)

\(= (a+d)(b-c)\)

Calculation:

Consider the polynomial \(m^{2} + 5m + mt + 5t\).

This is a four term polynomial, factorization of this polynomial can be find by factor by grouping as,

\(m^{2} + 5m + mt + 5t = (m^{2} + Sm) + (mt + 5t)\)

\(=m(m+5)+t(m+5)\)

As, \((m + 5)\) is the common factor of the polynomial factor it out as follows:

\(m^{2} + 5m + mt + 5t = (m + 5m) + (mt + 5t)\)

\(=m(m+5)+t(m+5)\)

\(=(m+5)(m+1)\)

The factorization of the polynomial \(m^{2} + 5m + mt + 5t\) is \((m+5)(m+t)\).

Check the result as follows:

\((m+5)(m+t})=m*m+m*t+5*m+5*t\)

\(=m^{2}+mt+5m+5t\)

\(=m^{2}+5m+mt+5t\)

Thus, the factorization of the polynomial \(m^{2} + 5m + mt + 5t\) is \((m+5)(m+t)\).

The factors of a polynomial can be find by taking a common factor and this method is called factor by grouping,

\(ab+ac+bd+cd=a(b+c)+d(b+c)\)

\(=(a+d)(b+c)\)

Or,

\(ab - ac + bd-cd = a(b-c)+d(b-c)\)

\(= (a+d)(b-c)\)

Calculation:

Consider the polynomial \(m^{2} + 5m + mt + 5t\).

This is a four term polynomial, factorization of this polynomial can be find by factor by grouping as,

\(m^{2} + 5m + mt + 5t = (m^{2} + Sm) + (mt + 5t)\)

\(=m(m+5)+t(m+5)\)

As, \((m + 5)\) is the common factor of the polynomial factor it out as follows:

\(m^{2} + 5m + mt + 5t = (m + 5m) + (mt + 5t)\)

\(=m(m+5)+t(m+5)\)

\(=(m+5)(m+1)\)

The factorization of the polynomial \(m^{2} + 5m + mt + 5t\) is \((m+5)(m+t)\).

Check the result as follows:

\((m+5)(m+t})=m*m+m*t+5*m+5*t\)

\(=m^{2}+mt+5m+5t\)

\(=m^{2}+5m+mt+5t\)

Thus, the factorization of the polynomial \(m^{2} + 5m + mt + 5t\) is \((m+5)(m+t)\).