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# Polynomial Factorization

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This problem was in my math textbook, and I found it really interesting. It took me a while to do, but the solution is kind of simple. If you want a fun little problem, here you go I geuss!

Good Luck!

The polynomial \(f(x) = 3x^2 - 20x^2+kx+12\) is divisible by \(x - 3\) for some constant k. Factor the polynomial f(x) completely.

I will leave a few hints.

First off, we must find what k is. Use synthetic division to your advantage.

Second, once we achieve k, factor x - 3 out of the polynomial to find the other 2 roots.

For bonus points, find the roots of the polynomial f(x).

Good Luck and Thank you!

:)

Apr 30, 2024

#1
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I think it must be  3x^3 - 20x^2 + kx + 12   ????

3 is one  root

3  [  3     -  20           k        12    ]

9        -33       3k - 99

____________________________

3       -11      k - 33     3k - 87

3k - 87 =  0

3k =  87

k = 29

3  [  3      -20         29       12  ]

9         -33     -12

_________________________

3        -11         -4         0

The remaining polynomial is

3x^2  -11x  - 4   = 0

(3x  + 1) ( x - 4)  = 0

The other two roots   are   -1/3   and   4

Apr 30, 2024
edited by CPhill  Apr 30, 2024
#3
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Good Job @CPhill and @MaxWong.

I geuss this problem wasn't much of a challenge for you. LOL!

Yes, I did the same thing as both of you. We can easily use synthetic division to achieve our eventual solution.

And yes, sorry about that. I did mean 3x^3. A little typo, but I'm glad you both realized it!

Thanks!

NotThatSmart  Apr 30, 2024
edited by NotThatSmart  May 1, 2024
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It's still a pretty good problem.....!!!

CPhill  Apr 30, 2024
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I assume you mean \(3x^3\) and not \(3x^2\), because it is impractical to write polynomials in an un-simplified way. (I hope you understand what I meant, my English is not that good)

We state the factor theorem as follows:

Let \(p(x)\) be a polynomial. If \(p(x)\) is divisible by \(x - k\), then \(p(k) = 0\).

This is a well-known result in algebra. We prove it using division algorithm: Suppose that when p(x) is divided by x - k, the quotient is q(x) and the remainder is r. (The remainder is a constant because it must have lower degree than the divisor x - k.) By division algorithm, we must have \(p(x) = (x - k)q(x) + r\), substituting x = k gives \(r = p(k) + 0q(k) = p(k)\). Therefore the factor theorem is equivalent to saying "If p(x) is divisible by x - k, then the remainder is 0 when p(x) is divided by by x - k", which is obviously true.

We can use this result to solve the problem. From the factor theorem, we know that \(f(3) = 0\). Then, substituting x = 3 into the definition of f(x) gives \(3k -87 = 0\), which gives k = 29.

Suppose \(f(x) = (x - 3)(ax^2 + bx + c)\). Then \(ax^3 + (b - 3a)x^2 + (c - 3b)x - 3c = 3x^2 - 20x^2 + 29x + 12\). Comparing coefficients gives \(a = 3, b = -11, c = -4\). Then we have \(f(x) = (x - 3)(3x^2 - 11x - 4)\). Factorizing the quadratic part further, we have \(f(x) = (x - 3)(3x + 1)(x - 4)\).

For the bonus part, simple inspection gives the roots x = 3, x = 4, and x = -1/3.

Apr 30, 2024