Your friend Chadwick is trying to factor the polynomial 6x3+8x2+15x+20
. "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.
Option 3: 5(2x2+3x)+4x(2x+5)
Option 3: 5(2x^2+3x)+4x(2x+5) is not equivalent to Chadwick’s polynomial.
To determine which option is not equivalent to Chadwick's polynomial, let's simplify each option and compare them to the original polynomial.
Option 1: 2x^2(3x+4) + 5(3x+4)
Using the distributive property, we have:
6x^3 + 8x^2 + 15x + 20
Option 2: 3x(2x^2+5) + 4(2x^2+5)
Using the distributive property, we have:
6x^3 + 15x + 8x^2 + 20
Option 3: 5(2x^2+3x) + 4x(2x+5)
Using the distributive property, we have:
10x^2 + 15x + 8x^2 + 20x + 20
Comparing the results, we can see that Option 2 matches Chadwick's original polynomial expression: 6x^3 + 8x^2 + 15x + 20. Therefore, Option 2 is equivalent to Chadwick's polynomial.
Option 1 and Option 3 are also equivalent to the original polynomial.
Thus, the answer is: Option 2 is not equivalent to Chadwick's polynomial.