Your friend Chadwick is trying to factor the polynomial 6x3+8x2+15x+20

Question

. "I noticed that I could group the first and the third terms; this would help me begin to factor the polynomial,” says Chadwick. You notice that another equally valid grouping would be to group the first and second terms, with a common factor of 2x2
, and the third and fourth terms, with a common factor of 5. Given that both of these first steps are valid, which of the following options is not equivalent to Chadwick’s polynomial?

Option 1: 2x2(3x+4)+5(3x+4)


Option 2: 3x(2x2+5)+4(2x2+5)


Option 3: 5(2x2+3x)+4x(2x+5)


(1 point)
Option #
is not equivalent to Chadwick’s polynomial.

Answer 1

Option 3: 5(2x2+3x)+4x(2x+5)