Answer:
[tex]\large\boxed{C.\ 2x^3-x^2-18x+9}[/tex]
Step-by-step explanation:
[tex]\text{Use}\ a^2-b^2=(a-b)(a+b)\qquad(*)\\\\\underbrace{(x-3)(x+3)}_{(*)}(2x-1)=(x^2-3^2)(2x-1)=(x^2-9)(2x-1)\\\\\text{Use FOIL:}\ (a+b)(c+d)=ac+ad+bc+bd\\\\=(x^2)(2x)+(x^2)(-1)+(-9)(2x)+(-9)(-1)\\\\=2x^3-x^2-18x+9[/tex]
Answer:
C
Step-by-step explanation:
Given
(x - 3)(x + 3)(2x - 1)
Expand (x - 3)(x + 3) = x² - 9 ← difference of squares factors, thus
= (x² - 9)(2x - 1)
Each term in the second factor is multiplied by each term in the first factor, that is
= x²(2x - 1) - 9(2x - 1) ← distribute both parenthesis
= 2x³ - x² - 18x + 9 → C
