The discussion focuses on manipulating the algebraic expression \(\frac{2y^{2}}{4+y^{2}}\) into the form \(2-\frac{8}{4+y^{2}}\). The transformation involves adding and subtracting 8 in the numerator, leading to the equation \(\frac{2y^2 + 8 - 8}{4+y^2} = \frac{2(4+y^2) - 8}{4+y^2}\). An alternative method presented is polynomial long division, which confirms the same result by dividing \(2y^2\) by \(y^2 + 4\) and finding a remainder of -8.
PREREQUISITESStudents studying calculus, educators teaching algebraic manipulation, and anyone looking to improve their skills in handling rational expressions.
cristo said:They appear to add 8 and subtract 8 from the numerator. That is,
\frac{2y^2}{4+y^2}=\frac{2y^2+8-8}{4+y^2}=\frac{2(4+y^2)-8}{4+y^2}=2-\frac{8}{4+y^2}
<br /> <br /> Thanks ;). Very clear to me now.HallsofIvy said:Another way to see that is simply "long division". y^2+ 0y+ 4 divides into 2y^2+ 0y+ 0 2 times with a remainder of 2y^2- 2(y^2+ 4)= 2y^2-2y^2- 8= -8[/tex] so<br /> \frac{2y^2}{y^2+ 4}= 2- \frac{8}{y^2+ 4}