Question

1. \\(\frac{21x^3 - 84x}{14x^2 - 35x}\\)

Explanation:

Step1: Factor numerator and denominator

Factor out common terms:
Numerator: \(21x^3 - 84x = 21x(x^2 - 4) = 21x(x - 2)(x + 2)\) (using difference of squares \(a^2 - b^2=(a - b)(a + b)\))
Denominator: \(14x^2 - 35x = 7x(2x - 5)\)

Wait, correction: Let's redo factoring properly.
Numerator: \(21x^3 - 84x = 21x(x^2 - 4)=21x(x - 2)(x + 2)\)? No, \(x^2 - 4\) is correct, but let's check GCF first. GCF of 21 and 84 is 21, GCF of \(x^3\) and \(x\) is \(x\), so \(21x(x^2 - 4)=21x(x - 2)(x + 2)\).
Denominator: \(14x^2 - 35x\), GCF of 14 and 35 is 7, GCF of \(x^2\) and \(x\) is \(x\), so \(7x(2x - 5)\). Wait, but 21 and 7 have GCF 7, x and x have GCF x. Wait, maybe better to factor step by step:

Wait, original numerator: \(21x^3 - 84x = 21x(x^2 - 4) = 21x(x - 2)(x + 2)\)
Denominator: \(14x^2 - 35x = 7x(2x - 5)\)

Wait, no, 21x and 7x: GCF is 7x. Let's factor numerator and denominator with GCF:

Numerator: \(21x^3 - 84x = 21x(x^2 - 4) = 3 \times 7x(x - 2)(x + 2)\)
Denominator: \(14x^2 - 35x = 7x(2x - 5) = 7x(2x - 5)\)

Now, cancel common factors: 7x is common? Wait, 21x = 3×7x, 7x is in denominator. So:

\(\frac{21x(x^2 - 4)}{14x^2 - 35x} = \frac{3 \times 7x(x - 2)(x + 2)}{7x(2x - 5)}\)

Cancel 7x (assuming \(x eq 0\) and \(2x - 5 eq 0\), \(x eq \frac{5}{2}\), \(x eq 0\)):

\(\frac{3(x - 2)(x + 2)}{2x - 5}\)? Wait, no, wait \(x^2 - 4\) is \((x - 2)(x + 2)\), but denominator after factoring is \(7x(2x - 5)\), numerator is \(21x(x^2 - 4)=21x(x - 2)(x + 2)\). So 21x /14x² -35x: let's factor numerator and denominator again:

Numerator: \(21x^3 - 84x = 21x(x^2 - 4) = 21x(x - 2)(x + 2)\)
Denominator: \(14x^2 - 35x = 7x(2x - 5)\)

Now, 21x and 7x: 21x = 3×7x, 7x is in denominator. So cancel 7x:

\(\frac{3(x - 2)(x + 2)}{2x - 5}\)? Wait, no, 21x /7x = 3, so:

\(\frac{3(x^2 - 4)}{2x - 5}\)? Wait, no, \(x^2 - 4\) is (x - 2)(x + 2), but denominator is 2x - 5. Wait, maybe I made a mistake in factoring. Let's try again:

Wait, original expression: \(\frac{21x^3 - 84x}{14x^2 - 35x}\)

Factor numerator: 21x³ -84x = 21x(x² - 4) = 21x(x - 2)(x + 2)
Factor denominator: 14x² -35x = 7x(2x - 5)

Now, 21x / 7x = 3, so:

\(\frac{3(x - 2)(x + 2)}{2x - 5}\)? Wait, no, 21x is 3*7x, 7x is in denominator, so cancel 7x:

\(\frac{3(x^2 - 4)}{2x - 5}\)? Wait, no, x² -4 is (x-2)(x+2), but denominator is 2x -5. Wait, maybe I messed up the denominator factoring. Let's factor denominator again: 14x² -35x = 7x(2x - 5). Correct. Numerator: 21x³ -84x = 21x(x² - 4) = 21x(x - 2)(x + 2). Correct.

Wait, but maybe another approach: factor numerator and denominator by GCF first:

Numerator: GCF of 21 and 84 is 21, GCF of x³ and x is x, so 21x(x² - 4)
Denominator: GCF of 14 and 35 is 7, GCF of x² and x is x, so 7x(2x - 5)

Now, 21x / 7x = 3, so:

\(\frac{3(x² - 4)}{2x - 5}\)? Wait, no, x² -4 is (x-2)(x+2), but denominator is 2x -5. Wait, maybe I made a mistake in the numerator's x² -4. Wait, 21x³ -84x: 21x³ -84x = 21x(x² - 4) = 21x(x - 2)(x + 2). Correct. Denominator: 14x² -35x = 7x(2x - 5). Correct.

Wait, but maybe the problem is to simplify the rational expression. So after canceling 7x (x ≠ 0), we have:

\(\frac{3(x² - 4)}{2x - 5}\)? Wait, no, 21x /7x is 3, so numerator becomes 3(x² - 4), denominator is (2x - 5). But x² -4 is (x - 2)(x + 2), so:

\(\frac{3(x - 2)(x + 2)}{2x - 5}\)

Wait, but maybe I made a mistake in factoring. Let's check with x=1:

Original expression: (21(1) -84(1))/(14(1) -35(1)) = (21 -84)/(14 -35) = (-63)/(-21) = 3

Simplified expression: 3(1 -2)(1 +2)/(2(1) -5) = 3(-1)(3)/(-3) = (-9)/(-3) = 3. Correct.

Another check: x=2 (but x=2: denominator 2(2)-5= -1, numerator 3(0)(4)/(-1)=0. Original expression: (21(8) -84(2))/(14(4) -35(2)) = (168 - 168)/(56 -70)=0/(-14)=0. Correct.

So the simplified form is \(\frac{3(x - 2)(x + 2)}{2x - 5}\) or \(\frac{3(x^2 - 4)}{2x - 5}\) or expanding numerator: \(\frac{3x^2 - 12}{2x - 5}\)

Wait, but maybe there's a simpler way. Let's factor numerator and denominator again:

Numerator: 21x³ -84x = 21x(x² - 4) = 21x(x - 2)(x + 2)
Denominator: 14x² -35x = 7x(2x - 5)

Cancel 7x from numerator (21x=3×7x) and denominator:

\(\frac{3(x - 2)(x + 2)}{2x - 5}\)

Yes, that's correct.

Answer:

\(\boxed{\dfrac{3(x - 2)(x + 2)}{2x - 5}}\) (or simplified as \(\dfrac{3x^2 - 12}{2x - 5}\))