Explanation
[tex]\dfrac{4e^2+16e-9}{2ef+12e-f-6}[/tex]
⇒ First, factor the numerator by grouping:
[tex]=\dfrac{4e^2-2e+18e-9}{2ef+12e-f-6}\\\\\\=\dfrac{2e(2e-1)+9(2e-1)}{2ef+12e-f-6}\\\\\\=\dfrac{(2e+9)(2e-1)}{2ef+12e-f-6}[/tex]
⇒ Now, factor the denominator by grouping:
[tex]=\dfrac{(2e+9)(2e-1)}{2e(f+6)-(f+6)}\\\\\\=\dfrac{(2e+9)(2e-1)}{(2e-1)(f+6)}[/tex]
⇒ We must determine which values of e and f are unacceptable, meaning, will make this expression undefined. These will be the values of e and f that make the denominator equal to 0.
- ⇒ To find these values, let's set each term in the denominator equal to 0, and solve for e and f.
- [tex]2e-1=0[/tex] ⇒ [tex]2e=1[/tex] ⇒ [tex]e=\dfrac{1}{2}[/tex]
- [tex]f+6=0[/tex] ⇒ [tex]f=-6[/tex]
- ⇒ The restrictions for e and f include [tex]e=\dfrac{1}{2}[/tex] and [tex]f=-6[/tex].
[tex]=\dfrac{(2e+9)(2e-1)}{(2e-1)(f+6)}[/tex]
⇒ Reduce values in the numerator and denominator:
[tex]=\dfrac{(2e+9)}{(f+6)}\\\\\\=\dfrac{2e+9}{f+6}[/tex]
Answer
[tex]=\dfrac{2e+9}{f+6}[/tex]
