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If the polynomial P(x) has factors of 12, (x - 5), and (x + 4), which of the following must also be a factor of P(x)?
  • a)
    2x2 + 8
  • b)
    4x2 - 20
  • c)
    6x2 - 6x - 120
  • d)
    x2 - 10x + 25
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If the polynomial P(x) has factors of 12, (x - 5), and (x + 4), which ...
Solution:

Given:
- The polynomial P(x) has factors of 12, (x - 5), and (x + 4).

To find:
- Which of the following must also be a factor of P(x).

Approach:
- We can find the remaining factor of P(x) by dividing P(x) by the given factors.
- If the remainder is zero, then the factor we divided by is indeed a factor of P(x).

Calculation:
1. Let's start by finding the first factor 12.
- Divide P(x) by 12 and check the remainder.

P(x) = 12(x - 5)(x + 4) (Given)
P(x)/12 = (x - 5)(x + 4)

To find the remainder, substitute x = 5 into the expression (x - 5)(x + 4):

(x - 5)(x + 4) = (5 - 5)(5 + 4) = 0

Since the remainder is zero, we can conclude that (x - 5)(x + 4) is a factor of P(x).

2. Now, let's move on to the second factor (x - 5).
- Divide P(x) by (x - 5) and check the remainder.

P(x)/(x - 5) = 12(x + 4)

To find the remainder, substitute x = 5 into the expression 12(x + 4):

12(x + 4) = 12(5 + 4) = 108

Since the remainder is not zero, we can conclude that (x - 5) is not a factor of P(x).

3. Finally, let's divide P(x) by (x + 4) and check the remainder.

P(x)/(x + 4) = 12(x - 5)

To find the remainder, substitute x = -4 into the expression 12(x - 5):

12(x - 5) = 12(-4 - 5) = -108

Since the remainder is not zero, we can conclude that (x + 4) is not a factor of P(x).

Therefore, the only factor of P(x) among the given options is 12(x - 5)(x + 4), which matches with option C: 6x^2 - 6x - 120.

Hence, the correct answer is option C: 6x^2 - 6x - 120.
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Community Answer
If the polynomial P(x) has factors of 12, (x - 5), and (x + 4), which ...
The simplest polynomial with factors of 12, (x - 5), and (x + 4) is P(x) = 12(x - 5)(x + 4). The completely factored form (including the prime factorization of the coefficient) of this polynomial is P(x) = (2)2 (3)(x - 5)(x + 4).
Now, we can look at the factored form of each choice:
(A) 2x2 + 8 = 2(x2 + 8) (x2 + 8 is not factorable over the reals, but it does equal (x - √8i)(x + √8i)
(B) 4x2 - 20 = 4(x2 - 5) = (2)2(x - √5)(x - √5)
(C) 6x2 - 6x - 120 = 6(x2 - x - 20) = (2)(3)(x - 5)(x + 4)
(D) x2 - 10x + 25 = (x - 5)(x - 5)
Notice that every polynomial in (A), (B), and (D) contains at least one factor that is NOT in the factored form of P(x). (In (D), the factor (x - 5) appears twice, but it appears only once in P(x).) Only choice (C) contains ONLY factors that appear in P(x), so it is the only choice that must be a factor of P(x).
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If the polynomial P(x) has factors of 12, (x - 5), and (x + 4), which of the following must also be a factor of P(x)?a)2x2 + 8b)4x2 - 20c)6x2 - 6x - 120d)x2 - 10x + 25Correct answer is option 'C'. Can you explain this answer?
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