We have the expression:
[tex]\frac{x^3+x^2-7}{x-3}[/tex]We can simplify it by dividing the numerator by the denominator (division of polynomials):
[tex]\begin{gathered} \frac{x^3+x^2-7}{x-3} \\ \frac{x^3-3x^2+3x^2+x^2-7}{x-3}=\frac{x^2(x-3)}{x-3}+\frac{4x^2-7}{x-3}=x^2+\frac{4x^2-7}{x-3} \\ x^2+\frac{4x^2-12x}{x-3}+\frac{12x-7}{x-3}=x^2+\frac{4x(x-3)}{x-3}+\frac{12x-7}{x-3}=x^2+4x+\frac{12x-7}{x-3} \\ x^2+4x+\frac{12x-36}{x-3}+\frac{36-7}{x-3}=x^2+4x+\frac{12(x-3)}{x-3}+\frac{29}{x-3} \\ x^2+4x+12+\frac{29}{x-3} \end{gathered}[/tex]The answer is: x²+4x+12+29/(x-3)
