Question

[Algebra 2] Find the quotient with the necessary restrictions.​

Answer 1

Answer:

The correct answer is:   3x² (4x - 1) / (x - 4) (x - 3) ∧ restriction x ≠ 3, x ≠ 4, x ≠ 0 and x ≠ 1/4

Step-by-step explanation:

Given:

((16x² - 8x + 1) / (x² - 7x + 12)) : ((20x² - 5x) / 15x³) =

dividing with one fraction is the same as multiplying with its reciprocal value

((16x² - 8x + 1) / (x² - 7x + 12)) · (15x³ / (20x² - 5x))

First we need to factorize both numerators and denominators

16x² - 8x + 1 = (4x - 1)²   This is square binomial

x² - 7x + 12 = x² - 4x - 3x + 12 = x (x - 4) - 3 (x - 4) = (x - 4) ( x - 3)

20x² - 5x = 5x (4x - 1)

(4x - 1)² / (x - 4) (x - 3) ·  15x³ / 5x (4x - 1)

The existence of this rational algebraic expression is possible only if it is:

x - 4 ≠ 0 and x - 3 ≠ 0 and x ≠ 0 and 4x - 1 ≠ 0  =>

x ≠ 4 and x ≠ 3 and x ≠ 0 and x ≠ 1/4  This is restriction

Finally we have:

3 x² (4x - 1) / (x - 4) (x - 3)

God with you!!!