[tex]4x-3,x\ne-5[/tex]
Explanation
[tex]\begin{gathered} f(x)=4x^2+17x-15 \\ g(x)=x+5 \end{gathered}[/tex]
find
[tex]\frac{f(x)}{g(x)}=\frac{4x^2+17x-15}{x+1}[/tex]
Step 1
factorize the numerator.
so
[tex]\begin{gathered} 4x^2+17x-15 \\ \text{rewrite 17x as}-3x+20x \\ 4x^2-3x+20x-15 \\ group \\ (4x^2-3x)+(20x-15) \\ x(4x-3)+5(4x-3) \\ \text{hence} \\ (x+5)(4x-3) \\ 4x^2+17x-15=(x+5)(4x-3) \end{gathered}[/tex]
Step 2
replace
[tex]\begin{gathered} \frac{f(x)}{g(x)}=\frac{4x^2+17x-15}{x+5} \\ \frac{f(x)}{g(x)}=\frac{(x+5)(4x-3)}{x+5} \\ (x+5)\text{ is eliminated,therefore} \\ \frac{f(x)}{g(x)}=(4x-3) \end{gathered}[/tex]
so, the answer is
[tex]4x-3[/tex]
but, we have the restriction, the denominator cannot be zero, so
[tex]\begin{gathered} \frac{f(x)}{g(x)}=\frac{(x+5)(4x-3)}{x+5} \\ x+5\ne0 \\ x\ne-5 \end{gathered}[/tex]
therefore, the solution is
[tex]4x-3,x\ne-5[/tex]
D)
I hope this helps you
