Factor \(x^2+9x+20\)

\(x^2+9x+20\)

Factor \(x^2+9x+20\)

=\((x+b)(x+c)\)

\(x^2+9x+20\)

Factor \(x^2+9x+20\)

=\((x+b)(x+c)\)

=\(x^2+x(b+c)+bc\)

\(x^2+9x+20\)

Factor \(x^2+9x+20\)

=\((x+b)(x+c)\)

=\(x^2+x(b+c)+bc\)

We must find \(b\) and \(c\) such that \(bc=20\) and \(b+c=9\). That is to say that \(b\) and \(c\) are factors of \(20\).

Factors of 20

\(1 \times 20\)

\(-1\times -20\)

\(2 \times 10\)

Factors of 20

\(1 \times 20\)

\(-1\times -20\)

\(2 \times 10\)

\(-2 \times -10\)

\(4 \times 5\)

\(-4\times -5\)

Factors of 20

\(1 \times 20\)

\(-1\times -20\)

\(2 \times 10\)

\(-2 \times -10\)

\(4 \times 5\)

\(-4\times -5\)

We are looking for two factors of \(20\) that add to \(9\)...\(4\) and \(5\) seem to work!

Factors of 20

Thus, \(x^2+9x+20\)=\((x+4)(x+5)\).

Factor \(x^2+9x+20\)

If we were to multiply out

\((x+4)(x+5)\), we would get

\(x^2+9x+20\). In other words,

\((x+4)\) and \((x+5)\) are factors of \(x^2+9x+20\).

Factor \(x^2+9x+20\)