To determine which binomial is a factor of the polynomial x³ – 4x² + x – 6, we can use the factor theorem or perform polynomial division.
The polynomial can be factored by checking potential roots. Possible rational roots can be derived from the factors of the constant term (-6) divided by the leading coefficient (1). Thus, the potential roots can be:
- ±1
- ±2
- ±3
- ±6
After testing these values, we find:
- Plugging in x = 1:
- x³ – 4x² + x – 6 = 1 – 4 + 1 – 6 = -8 (not a factor)
- Plugging in x = -1:
- x³ – 4x² + x – 6 = -1 – 4 – 1 – 6 = -12 (not a factor)
- Plugging in x = 2:
- x³ – 4x² + x – 6 = 8 – 16 + 2 – 6 = -12 (not a factor)
- Plugging in x = -2:
- x³ – 4x² + x – 6 = -8 – 16 – 2 – 6 = -32 (not a factor)
- Plugging in x = 3:
- x³ – 4x² + x – 6 = 27 – 36 + 3 – 6 = -12 (not a factor)
- Plugging in x = -3:
- x³ – 4x² + x – 6 = -27 – 36 – 3 – 6 = -72 (not a factor)
- Plugging in x = 6:
- x³ – 4x² + x – 6 = 216 – 144 + 6 – 6 = 72 (not a factor)
- Plugging in x = -6:
- x³ – 4x² + x – 6 = -216 – 144 – 6 – 6 = -372 (not a factor)
Since simple trials do not yield integer roots, we can try grouping terms or using synthetic division. However, if you are looking for specific binomials, you would typically pre-determine a set from a list of choices. Let’s assume we need to check the binomial factors like:
- (x – 2)
- (x + 2)
- (x – 3)
- (x + 3)
If we divide x³ – 4x² + x – 6 by (x – 2), using synthetic or long division, we may find a quotient showing whether this binomial is a factor. Achieving zero remainder would imply that (x – 2) is indeed a factor.
In summary, to establish which of the proposed binomials is truly a factor, proceed with testing each binomial via substitution or polynomial division. From observation, if (x – 2) yields a zero remainder, it is confirmed as a factor of the polynomial x³ – 4x² + x – 6.