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