Jiaern0405idn
Answered

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[tex] ({4e }^{2} + 16e - 9) \div (2ef + 12e - f - 6)[/tex]
please help me solve this problem​

Sagot :

Theleangreenbeanidn

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]

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