Question

%20-%206%29" id="TexFormula1" title=" ({4e }^{2} + 16e - 9) \div (2ef + 12e - f - 6)" alt=" ({4e }^{2} + 16e - 9) \div (2ef + 12e - f - 6)" align="absmiddle" class="latex-formula">
please help me solve this problem​

Answer 1

Explanation

$\dfrac{4e^2+16e-9}{2ef+12e-f-6}$

⇒ First, factor the numerator by grouping:

$=\dfrac{4e^2-2e+18e-9}{2ef+12e-f-6}\\\\\\=\dfrac{2e(2e-1)+9(2e-1)}{2ef+12e-f-6}\\\\\\=\dfrac{(2e+9)(2e-1)}{2ef+12e-f-6}$

⇒ Now, factor the denominator by grouping:

$=\dfrac{(2e+9)(2e-1)}{2e(f+6)-(f+6)}\\\\\\=\dfrac{(2e+9)(2e-1)}{(2e-1)(f+6)}$

⇒ 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.
• $2e-1=0$ ⇒ $2e=1$ ⇒ $e=\dfrac{1}{2}$
• $f+6=0$ ⇒ $f=-6$
• ⇒ The restrictions for e and f include $e=\dfrac{1}{2}$ and $f=-6$.

$=\dfrac{(2e+9)(2e-1)}{(2e-1)(f+6)}$

⇒ Reduce values in the numerator and denominator:

$=\dfrac{(2e+9)}{(f+6)}\\\\\\=\dfrac{2e+9}{f+6}$

Answer

$=\dfrac{2e+9}{f+6}$