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Problem statement:

Find a monic quadratic polynomial, $f(x)$, which divides both $$g(x) = 12x^3 − 30x^2 + 18x − 12$$ and $$h(x) = 6x^4 + 3x^3 + 6x^2 + 3$$

My take on it:

I divided $h(x)$ by $g(x)$ to get the quotient and remainder such that $$6x^4 + 3x^3 + 6x^2 + 3 = (12x^3 − 30x^2 + 18x − 12)(\frac{1}{2}x + \frac{3}{2}) + 3(14x^2 - 11x + 7)$$

It is also the case that any polynomial divisor of both $g(x)$ and $h(x)$ must also divide the remainder polynomial when $h(x)$ is divided by $g(x)$.

So following on from that, our common factor for $g(x)$ and $h(x)$ that we're trying to find, would also have to be a factor of our remainder, $3(14x^2 - 11x + 7)$. Yet the remainder cannot be factorised any further to turn it into a monic quadratic polynomial.

How should I go about solving this problem?

So far, it's only this particular problem that I find myself unable to solve, other ones like it are fine.

Important: Please don't provide the full solution but rather only the way forward as I don't want to be accused of plagiarism. This exact problem is part of an assignment.


EDIT: I actually had an error in my calculation, the resulting equation is now as follows $$ 6x^4 + 3x^3 + 6x^2 + 3 = (12x^3 − 30x^2 + 18x − 12)(\frac{1}{2}x + \frac{3}{2}) + 42(x^2 - \frac{1}{2}x + \frac{1}{2}) $$

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    $\begingroup$ I think you made an error in your computations as Alpha shows the remainder to be $3(10x^3-20x^2+11x-7)$. Your approach is a good one. Now find a linear combination of $h(x)$ and the remainder that removes the cubic term and you should be left with a multiple of the desired quadratic. $\endgroup$ Oct 1, 2017 at 15:43
  • $\begingroup$ @RossMillikan Thanks! I actually had a typo in my question (- instead of +). But I also had a calculation error as you suggested. Now I get $42(x^2 - \frac{1}{2}x + \frac{1}{2}$, which is basically what the other answer suggested. $\endgroup$ Oct 1, 2017 at 16:18

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we have: $$ 12x^3-30x^2+18x-12=6(2x^3-5x^2+3x-2) $$

and we see that $x=2$ is a root, so, dividing by $(x-2)$ we find: $$ 12x^3-30x^2+18x-12=6(x-2)(2x^2-x+1) $$

Now show that $2x^2-x+1$ divides the other polynomial. And you can write it as a monic polynomial using: $$ 2\left(x^2-\frac{1}{2}x+\frac{1}{2} \right) $$

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    $\begingroup$ The required polynomial must be monic, though. $\endgroup$ Oct 1, 2017 at 15:23
  • $\begingroup$ @GTonyJacobs : added to the answer. $\endgroup$ Oct 1, 2017 at 15:38
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$$ \left( 6 x^{4} + 3 x^{3} + 6 x^{2} + 3 \right) $$

$$ \left( 12 x^{3} - 30 x^{2} + 18 x - 12 \right) $$

$$ \left( 6 x^{4} + 3 x^{3} + 6 x^{2} + 3 \right) = \left( 12 x^{3} - 30 x^{2} + 18 x - 12 \right) \cdot \color{magenta}{ \left( \frac{ x + 3 }{ 2 } \right) } + \left( 42 x^{2} - 21 x + 21 \right) $$ $$ \left( 12 x^{3} - 30 x^{2} + 18 x - 12 \right) = \left( 42 x^{2} - 21 x + 21 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 4 }{ 7 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( \frac{ x + 3 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x + 3 }{ 2 } \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ 2 x - 4 }{ 7 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{2} + x + 1 }{ 7 } \right) }{ \left( \frac{ 2 x - 4 }{ 7 } \right) } $$ $$ \left( x^{2} + x + 1 \right) \left( \frac{ 1}{7 } \right) - \left( x - 2 \right) \left( \frac{ x + 3 }{ 7 } \right) = \left( 1 \right) $$ $$ \left( 6 x^{4} + 3 x^{3} + 6 x^{2} + 3 \right) = \left( x^{2} + x + 1 \right) \cdot \color{magenta}{ \left( 6 x^{2} - 3 x + 3 \right) } + \left( 0 \right) $$ $$ \left( 12 x^{3} - 30 x^{2} + 18 x - 12 \right) = \left( x - 2 \right) \cdot \color{magenta}{ \left( 6 x^{2} - 3 x + 3 \right) } + \left( 6 x^{3} - 15 x^{2} + 9 x - 6 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( 6 x^{2} - 3 x + 3 \right) } $$ $$ \left( 6 x^{4} + 3 x^{3} + 6 x^{2} + 3 \right) \left( \frac{ 1}{7 } \right) - \left( 6 x^{3} - 15 x^{2} + 9 x - 6 \right) \left( \frac{ x + 3 }{ 7 } \right) = \left( 6 x^{2} - 3 x + 3 \right) $$

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