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$f(x) = 2x^3 - 5x^2 + 7x + 10$

Given that $2x - 3$ is a factor, solve $f(x) = 0$ completely.

I have tried using a division of polynomials method to give a quadratic, but this gave a remainder of $16$. I have also used online calculators which suggest that there is no simple solution. Is there an error or should I be using a different method?

Thanks in advance.

Dando18
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user464136
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    but $2x-3$ is not a factor of $f(x)$ – Dr. Sonnhard Graubner Jul 17 '17 at 14:32
  • The solutions are not trivial and $2x-3$ is not a factor (http://www.wolframalpha.com/input/?i=2x%5E3%E2%88%925x%5E2%2B7*x%2B10%3D0). You can use Cardano's formula to determine the solutions (https://en.wikipedia.org/wiki/Cubic_function). – MrYouMath Jul 17 '17 at 14:32
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    There's an error. If $2x-3$ is a factor, then $3/2$ is a root. If you plug $3/2$ into the polynomial, you won't get $0$. – B. Goddard Jul 17 '17 at 14:33
  • have you checked that 3/2 is a root – hamam_Abdallah Jul 17 '17 at 14:34
  • $2x-3$ cannot be a factor, if it were than at $x=0$ we would have $3$ as a divisor of $10$ :-( . – Oscar Lanzi Jul 17 '17 at 14:36
  • the only real factor is $$\frac 1 6 \left(5 - \frac{17}{\sqrt[3]{3 \sqrt{59\ 757} - 730}} + \sqrt[3]{3 \sqrt{59\ 757} - 730}\right) $$ – Dando18 Jul 17 '17 at 14:37
  • Since $f\in \mathbb{Z}_3[X]$ has no root in $\mathbb{Z}_3$, it is irreducible in $\mathbb{Z}[X]$ and also in $\mathbb{Q}[X]$ by Gauß-Lemma, thus all roots are complex with at least one real root by the mean value theorem. If you want to solve $f(x)=0$, you can apply Cardano's method. – Fakemistake Jul 17 '17 at 15:01
  • I suspect there is an error somewhere. Maybe in copying the problem to MSE or in the source document containing the problem. – Χpẘ Jul 17 '17 at 15:33
  • $x-1$ is a factor of $f(x) = 2x^3 - 5x^2 - 7x + 10$. The quotient $\frac{2x^3 - 5x^2 + 7x + 10}{x-1}=2x^2-3x-10$ is somewhat similar to the alleged factor ($2x-3$) in the problem. Maybe the problem source document author lost their train of thought or something while writing the problem. – Χpẘ Jul 17 '17 at 15:48
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    Try with $2 x^3-5 x^2+7 x-6=0$ – Raffaele Jul 17 '17 at 18:50

2 Answers2

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If we put $x=\frac 32$ into $f(x)$ then we can see that \begin{align}f\left(\frac 32\right)&=2\left(\frac 32\right)^3−5\left(\frac 32\right)^2+7\left(\frac 32\right)+10\\ &=\frac {27}4-\frac {45}4+\frac{21}2+10\\ &=16\neq 0\end{align}

As $f\left(\frac 32\right)$ is not equal to $0$, then $2x-3$ is not a root.

This appears to be a homework/exam style question, are you sure you've written down $f(x)$ correctly? Either that, or the person who wrote the question mis-wrote $f(x)$ in the first place.

lioness99a
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Like many users in the comments I have entered the question into Wolfram Alpha and it did indeed give two complex roots and one real root, albeit not the $\frac{3}{2}$ that would be suggested by using factor theorem if the source question had been correct. It does look like the source question was flawed. Thanks for all the support.

Link to Wolfram Alpha solution page

exact

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dantopa
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user464136
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