we have
\(\frac{3x^{2}+8x-3}{x+3}\)
Simplify the numerator------> complete the square
equate the numerator to zero
\(3x^{2}+8x-3=0\)
Group terms that contain the same variable, and move the constant to the opposite side of the equation
\(3x^{2}+8x=3\)
Factor the leading coefficient
\(3(x^{2}+(8x/3))=3\)
Complete the square. Remember to balance the equation by adding the same constants to each side.
\(3(x^{2}+(8x/3)+(16/9))=3+(16/3)\)
\(3(x^{2}+(8x/3)+(16/9))=(25/3)\)
\((x^{2}+(8x/3)+(16/9))=(25/9)\)
Rewrite as perfect squares
\((x+(4/3))^{2}=(25/9)\)
Square root both sides
\(x+\frac{4}{3} =(+/-)\frac{5}{3}\)
\(x=-\frac{4}{3}(+/-)\frac{5}{3}\)
\(x=-\frac{4}{3}+\frac{5}{3}=\frac{1}{3}\)
\(x=-\frac{4}{3}-\frac{5}{3}=-3\)
so
\(3x^{2}+8x-3=3(x-\frac{1}{3})(x+3)=(3x-1)(x+3)\)
substitute
\(\frac{3x^{2}+8x-3}{x+3}=\frac{(3x-1)(x+3)}{x+3}=(3x-1)\)
therefore
the answer is the option B
\((3x-1)\)
