SOLUTION: Let P(x) be a polynomial of the form P(x) = 2x^3 + ax^2 - 23x + c, such that 12 and 7 are roots of P(x). What is the third root? For the polynomial in part (a), compute the or

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Question 1209715: Let P(x) be a polynomial of the form
P(x) = 2x^3 + ax^2 - 23x + c,
such that 12 and 7 are roots of P(x). What is the third root?
For the polynomial in part (a), compute the ordered pair (a,c).
Found 4 solutions by CPhill, greenestamps, ikleyn, mccravyedwin:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website!
Here's how to find the third root and the values of *a* and *c*:
1. **Use the factor theorem:** Since 1/2 and 7 are roots of P(x), then (x - 1/2) and (x - 7) are factors of P(x). Therefore, we can write P(x) as:
P(x) = 2(x - 1/2)(x - 7)(x - r)
where 'r' is the third root. The factor of 2 is included so that the coefficient of the x³ term matches the given polynomial.
2. **Expand the factored form:**
P(x) = 2(x - 1/2)(x - 7)(x - r)
P(x) = (2x - 1)(x - 7)(x - r)
P(x) = (2x² - 15x + 7)(x - r)
P(x) = 2x³ - 15x² + 7x - 2rx² + 15rx - 7r
P(x) = 2x³ + (-15 - 2r)x² + (7 + 15r)x - 7r
3. **Compare coefficients:** Now, compare the coefficients of the expanded form with the given form P(x) = 2x³ + ax² - 23x + c:
* Coefficient of x²: a = -15 - 2r
* Coefficient of x: -23 = 7 + 15r
* Constant term: c = -7r
4. **Solve for r:** From the coefficient of x, we can solve for r:
-23 = 7 + 15r
-30 = 15r
r = -2
5. **Find a and c:** Now that we know r = -2, we can find *a* and *c*:
a = -15 - 2(-2) = -15 + 4 = -11
c = -7(-2) = 14
Therefore, the third root is -2, and the ordered pair (a, c) is (-11, 14).

Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!

The AI solution from the other tutor is wrong from the first line, since it uses roots 1/2 and 7 instead of 12 and 7.

The process used, however, is fine.

Follow the process shown in that solution using the correct two given roots to find....

third root: -191/38

a = -531/19

c = 16044/19

Then you can check those results by graphing the polynomial using desmos.com

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NOTE: it is possible that the original problem somehow got corrupted, and that the correct give roots were in fact 1/2 and 7. I say that because those roots give much "nicer" answers for the third root (and for the coefficients a and c).

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While that path to the answer is fine, another path is perhaps easier, requiring arithmetic that is not quite as nasty.

7 and 12 are roots:

7:

[1]

12:

[2]

Comparing [1] and [2]...

Then

So (a,c) = (-531/19,16044/19)

Find the third root knowing that the sum of the roots is -a/2.

Answer by ikleyn(52851)   (Show Source):
You can put this solution on YOUR website!
.
Let P(x) be a polynomial of the form
P(x) = 2x^3 + ax^2 - 23x + c,
such that 12 and 7 are roots of P(x). What is the third root?
For the polynomial in part (a), compute the ordered pair (a,c).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @CPhill is INCORRECT.

        It is incorrect, since he mistakenly uses 1/2 instead of 12 in all his calculations through his post.

Answer by mccravyedwin(408)   (Show Source):
You can put this solution on YOUR website!

Go to https://www.wolframalpha.com/ Type in: 2*7^3 + a*7^2-23*7 + c = 0, 2*12^3 + a*12^2 - 23*12 + c = 0 Press ENTER Get a = -531/19, c= 16044/19. So you have the second part of the problem (a,c) = (-531/19, 16044/19) Now erase what you typed in before. Type in: 2x^3 + ax^2 - 23x + c = 0, a = -531/19, c= 16044/19 Press ENTER See the third root: x = -191/38 Edwin