(p - 1)(p - 2)(p - 3) all over p+3 multiply p² - 9 all over p² - 3p + 2

1 answer

First, factor the numerator:

(p - 1)(p - 2)(p - 3) = p³ - 6p² + 11p - 6

Then, factor the denominator:

p² - 3p + 2 = (p - 1)(p - 2)

(p + 3) cancels out from both numerator and denominator, leaving:

(p³ - 6p² + 11p - 6) ÷ (p - 1)(p - 2)

Now we can use partial fraction decomposition to separate the fraction into simpler parts:

(p³ - 6p² + 11p - 6) ÷ (p - 1)(p - 2) = A ÷ (p - 1) + B ÷ (p - 2)

To solve for A and B, we can use common denominator:

A(p - 2) + B(p - 1) = p³ - 6p² + 11p - 6

Setting p = 1, we get:

A(-1) + B(0) = -4

A = 4

Setting p = 2, we get:

A(0) + B(1) = 2

B = 2

Therefore, the original fraction can be rewritten as:

4 ÷ (p - 1) + 2 ÷ (p - 2)

Simplifying, we get:

4 ÷ (p - 1) + 2 ÷ (p - 2) = (4p - 6) ÷ (p² - 3p + 2)

Finally, we can substitute this expression back into the original problem:

(p - 1)(p - 2)(p - 3) ÷ (p + 3) × (p² - 9) ÷ (p² - 3p + 2)

= (p - 1)(p - 2)(p - 3) ÷ (p + 3) × [(p - 3)(p + 3) ÷ (p - 1)(p - 2)]

= (p - 1)(p - 2)(p - 3) ÷ (p - 1)(p - 2)

= p - 3

Therefore, the final answer is p - 3.