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!

:)

NotThatSmart Apr 30, 2024

#1**+1 **

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

CPhill Apr 30, 2024

#3**+1 **

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

#2**0 **

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.

MaxWong Apr 30, 2024