Answer:
[tex]f(x)\div g(x)=-10[/tex]
Step-by-step explanation:
Solve [tex]f(x)[/tex]
[tex]f(x)=x^2-19x+90[/tex]
[tex]x^2-19x+90[/tex]
[tex]\mathrm{Break\:the\:expression\:into\:groups}[/tex]
[tex]\left(x^2-9x\right)+\left(-10x+90\right)[/tex]
[tex]\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2-9x[/tex]
[tex]=x(x-9)[/tex]
[tex]\mathrm{Factor\:out\:-10\;from}\:\:-10x+90[/tex]
[tex]=-10(x-9)[/tex]
[tex]=x\left(x-9\right)-10\left(x-9\right)[/tex]
[tex]\mathrm{Factor\:out\:common\:term\:}x-9[/tex]
[tex]\bold{\left(x-9\right)\left(x-10\right)}[/tex]
Solve [tex]g(x)[/tex]
[tex]g(x)=x-9[/tex]
[tex]\bold{x-9}[/tex]
[tex](x-9\right)\left(x-10)}\div \:x-9[/tex]
Now, solve:
[tex](x-9\right)\left(x-10)}\div \:x-9[/tex]
[tex]=\frac{(x-9)\left(x-10)}{:x-9}[/tex]
[tex]= -10[/tex]
Therefore, [tex]f(x)\div g(x)=-10[/tex]
