Question

question 9 of 10
which of the following is the correct factorization of the polynomial below?
$2p^2 - 10pq + 25q^2$

a. $(2p - 5q)(p - 5q)$
b. $(2p - 5q)(p + 5q)$
c. $(2p - 5q)(2p^2 + 2q)$
d. the polynomial is irreducible.

Explanation:

Step1: Expand Option A

Multiply \((2p - 5q)(p - 5q)\) using the distributive property (FOIL method):
\(2p \cdot p + 2p \cdot (-5q) - 5q \cdot p + (-5q) \cdot (-5q) = 2p^2 - 10pq - 5pq + 25q^2 = 2p^2 - 15pq + 25q^2\). This does not match the original polynomial \(2p^2 - 10pq + 25q^2\).

Step2: Expand Option B

Multiply \((2p - 5q)(p + 5q)\):
\(2p \cdot p + 2p \cdot 5q - 5q \cdot p + (-5q) \cdot 5q = 2p^2 + 10pq - 5pq - 25q^2 = 2p^2 + 5pq - 25q^2\). This does not match the original polynomial.

Step3: Expand Option C

Multiply \((2p - 5q)(2p^2 + 2q)\):
\(2p \cdot 2p^2 + 2p \cdot 2q - 5q \cdot 2p^2 - 5q \cdot 2q = 4p^3 + 4pq - 10p^2q - 10q^2\). This is a cubic polynomial, while the original is quadratic, so it cannot match.

Step4: Analyze Irreducibility

Since the polynomial is a quadratic in two variables (\(p\) and \(q\)), and attempts to factor it into linear terms (as in options A, B, C) fail (as shown in steps 1–3), the polynomial is irreducible over the set of polynomials with integer coefficients.

Answer:

D. The polynomial is irreducible.