Answer:
3x-1
Step-by-step explanation:
We have the following expression:
[tex]\frac{3x^2+8x-3}{x+3}[/tex]
Equating this equal to zero and completing the square then to get:
[tex]3x^2+8x-3=0[/tex]
Arranging the terms which contain the same variable and moving the constant to the opposite side of the equation:
[tex]3x^2+8x=3[/tex]
Factorize the leading coefficient:
[tex]3(x^2+\frac{8x}{3} )=3[/tex]
Now complete the square and do not forget to balance the equation by adding the same constants to each side.
[tex]3(x^2+\frac{8x}{3} +\frac{16}{9} )=3+\frac{16}{3}[/tex]
[tex]3(x^2+\frac{8x}{3} +\frac{16}{9} )=\frac{25}{3}[/tex]
[tex](x^2+\frac{8x}{3} +\frac{16}{9} )=\frac{25}{9}[/tex]
Re-writing it as perfect square:
[tex](x+\frac{4}{3} )^2=\frac{25}{9}[/tex]
Taking square root at both sides to get:
[tex]x+\frac{4}{3} =+-\frac{5}{3}[/tex]
[tex]x=-\frac{4}{3} +\frac{5}{3} ,x=-\frac{4}{3} -\frac{5}{3}[/tex]
[tex]x=\frac{1}{3} , x=-3[/tex]
So, [tex]3x^2+8x-3=3(x-\frac{1}{3})(x+3) =(3x-1)(x+3)[/tex]
Substiting it:
[tex]\frac{3x^2+8x-3}{x+3}[/tex] = [tex]\frac{(3x-1)(x+3)}{(x+3)} = (3x-1)[/tex]
