Algebra HELP?

Question

I tried to do these but I keep getting confused, please help.

1) Find the LCM of 9-4x^2, 3-2x, and 3 + 2x

2)18/ 5x^3 multiplied by 5x^2/6

3)x + 2/( x+3)(x-5)(x-1)

4) c-3 divided by 3-c

Answer 1

1).

First factor each component :

9 - 4x² = 3² - (2x)² = (3 + 2x)(3 - 2x)

this means that, (9 - 4x²) is divisible by both (3 + 2x) and (3 - 2x)

(3 - 2x) is prime, and

(3 + 2x) is prime.

The Least Common Multiples of 9 - 4x² , 3 - 2x, and, 3 + 2x is product of (3 - 2x) and (3 + 2x).

Therefore, the LCM of 9 - 4x² , 3 - 2x, and, 3 + 2x is 9 - 4x² .

2).

(18/5x³) * (5x²/6)

= ( 18 * 5x²)/(5x³ * 6)

= ( 6 * 3 * 5x²)/(5x² * x * 6)

cancel common terms.

= 3/x.

Therefore, (18/5x³) * (5x²/6) = 3/x.

3).

x + {2/[(x + 3)(x - 5)(x - 1)]}

LCM of [(x + 3)(x - 5)(x - 1)] and 1 is [(x + 3)(x - 5)(x - 1)].

= {x[(x + 3)(x - 5)(x - 1)] + 2}/[(x + 3)(x - 5)(x - 1)]

= {x[x³ - 3x² - 13x + 15] + 2}/[(x + 3)(x - 5)(x - 1)]

= {x⁴ - 3x³ - 13x² + 15x + 2}/[(x + 3)(x - 5)(x - 1)].

4).

(c - 3)/(3 - c)

Common out negative 1 from the denominator.

= - (c - 3)/(c - 3)

Cancel common terms.

= - 1.

Therefore, (c - 3)/(3 - c) = - 1.