Question

which of the following is the correct factorization of the polynomial below? 3p^2 - 15pq + 20q^2
a. (3p - 5q)(p - 10q)
b. (3p - 5q)(p + 5q)
c. (3p - 5q)(2p^2 + 5q)
d. the polynomial is irreducible.

Explanation:

Step1: Check for common factors

There is no common factor among the terms $3p^{2}$, $- 15pq$ and $20q^{2}$ other than 1.

Step2: Try factoring as a quadratic in terms of $p$

For a quadratic polynomial $ax^{2}+bx + c$ (here $x = p$, $a = 3$, $b=-15q$, $c = 20q^{2}$), we can use the quadratic - formula $p=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Calculate the discriminant $\Delta=b^{2}-4ac=(-15q)^{2}-4\times3\times20q^{2}=225q^{2}-240q^{2}=- 15q^{2}<0$ when considering it as a quadratic in $p$.

Step3: Conclusion

Since the discriminant is negative and there are no simple binomial factors by inspection, the polynomial $3p^{2}-15pq + 20q^{2}$ is irreducible over the real - numbers.

Answer:

D. The polynomial is irreducible.