Answer:
The answer is \((3x+1)\)
Step-by-step explanation:
we have
\(\frac{3x^{2}-11x-4}{x-4}\)
Convert the numerator in factored form
\(3x^{2}-11x-4=0\)
Group terms that contain the same variable, and move the constant to the opposite side of the equation
\(3x^{2}-11x=4\)
Factor the leading coefficient
\(3(x^{2}-11x/3)=4\)
Complete the square. Remember to balance the equation by adding the same constants to each side.
\(3(x^{2}-11x/3+(121/36))=4+(121/12)\)
\(3(x^{2}-11x/3+(121/36))=(169/12)\)
\((x^{2}-11x/3+(121/36))=(169/36)\)
Rewrite as perfect squares
\((x-(11/6))^{2}=(169/36)\)
Square root both sides
\(x-\frac{11}{6}=(+/-)\frac{13}{6}\)
\(x=\frac{11}{6}(+/-)\frac{13}{6}\)
\(x=\frac{11}{6}+\frac{13}{6}=4\)
\(x=\frac{11}{6}-\frac{13}{6}=-\frac{1}{3}\)
therefore
\(3x^{2}-11x-4=3(x-4)(x+\frac{1}{3})=(x-4)(3x+1)\)
Substitute
\(\frac{3x^{2}-11x-4}{x-4}=\frac{(x-4)(3x+1)}{x-4}=(3x+1)\)
