Formula used:

The factors of a polynomial can be found 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 \(a^{2}+ac+a+c\).

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

\(a^{2}+ac+a+c=(a^{2}+ac)+(a+c)\)

\(=a(a+c)+(a+c)\)

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

\(a^{2}+ac+a+c=a(a+c)+(a+c)\)

\(=(a+c)(a+1)\)

The factorization of the polynomial \(a^{2}+ac+a+c\) is \((a+c)(a+ 1)\).

Check the result as follows:

\((a+c)(a+1)=a*a+a*1+c*a+c*1\)

\(=a^{2}+ac+a+c\)

Thus, the factorization of the polynomial \(a^{2}+ac+a+c\) is \((a+c)(a+1)\).

The factors of a polynomial can be found 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 \(a^{2}+ac+a+c\).

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

\(a^{2}+ac+a+c=(a^{2}+ac)+(a+c)\)

\(=a(a+c)+(a+c)\)

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

\(a^{2}+ac+a+c=a(a+c)+(a+c)\)

\(=(a+c)(a+1)\)

The factorization of the polynomial \(a^{2}+ac+a+c\) is \((a+c)(a+ 1)\).

Check the result as follows:

\((a+c)(a+1)=a*a+a*1+c*a+c*1\)

\(=a^{2}+ac+a+c\)

Thus, the factorization of the polynomial \(a^{2}+ac+a+c\) is \((a+c)(a+1)\).