Solve the following quadratic expressions by factoring. First write the expressions in completely factored form....

1) Given: ${\displaystyle x^2+9x+20=0}$
${\displaystyle \Rightarrow x^2+4x+5x+20=0}$
${\displaystyle \Rightarrow x(x+4)+5(x+4)=0}$
${\displaystyle \Rightarrow (x+4)(x+5)=0}$
Factored form
$(x+4)(x+5)$Real number solution
${\displaystyle \Rightarrow (x+4)(x+5)=0}$
${\displaystyle \Rightarrow x=-4,x=-5}$
${\displaystyle x=-4,-5}$
2) ${\displaystyle \Rightarrow x^2-x-72=0}$
${\displaystyle \Rightarrow x^2-9x+8x-72=0}$
${\displaystyle \Rightarrow x(x-9)+8(x-9)=0}$
${\displaystyle \Rightarrow (x-9)(x+8)}$
Factored form
${\displaystyle (x-9)(x+8)}$
Real number solution
${\displaystyle \Rightarrow (x-9)(x+8)=0}$
${\displaystyle \Rightarrow x=9,x=-8}$
${\displaystyle x=9,-8}$
3) ${\displaystyle x^2-36=0}$
${\displaystyle \Rightarrow x^2=36}$
${\displaystyle \Rightarrow x=\sqrt{36}}$
${\displaystyle \Rightarrow x=6}$
Factored form.
${\displaystyle x=6}$
Real number solution.
${\displaystyle x=0.6}$
4) ${\displaystyle x^2-10x+25=0}$
${\displaystyle \Rightarrow x^2-5x-5x+25=0}$
${\displaystyle \Rightarrow x^2-5x-5x+25=0}$
${\displaystyle \Rightarrow x(x-5)-5(x-5)=0}$
${\displaystyle \Rightarrow (x-5)(x-5)=0}$
Factored form
${\displaystyle (x-5)(x-5)}$
Real number solution
${\displaystyle (x-5)(x-5)=0}$
${\displaystyle x=5,5}$