Question

Factorize:

$9a^2+\frac{1}{9a^2}-2-12a+\frac{4}{3a}$.

• A. $(3a-\frac{1}{3a})(3a-\frac{1}{3a}-4)$
• B. $(a-\frac{1}{3a})(7a-\frac{1}{3a}-1)$
• C. $(3a-\frac{1}{3a})(2a-\frac{1}{3a}-4)$
• D. $(3a-\frac{1}{3a})(4a-\frac{1}{3a}-1)$

Answer 1

Answer: (A)

$(3a-\frac{1}{3a})(3a-\frac{1}{3a}-4)$

$9a^2+\frac{1}{9a^2}-2-12a+\frac{4}{3a}$

$=(3a{)}^2+(\frac{1}{3a}{)}^2-2\left(3a\right)\left(\frac{1}{3a}\right)-4(3a-\frac{1}{3a})$

$=(3a-\frac{1}{3a}{)}^2-4(3a-\frac{1}{3a})$

$=(3a-\frac{1}{3a})(3a-\frac{1}{3a}-4)$