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Determine the number of real roots in the equation $2x^3 + x^2 = 3$.

I know about finding the different roots, and solving giving that it has (for example) 2i as a root, but I'm not sure how to just find the amount of real roots. I've found nothing that can help so far.

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  • $\begingroup$ Welcome to math.SE! Please provide your work and thoughts about this problem. $\endgroup$ – Don Larynx Sep 21 '13 at 13:28
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    $\begingroup$ @Jossie This seems to be the standard invitation to the newcomers here? ;) $\endgroup$ – W_D Sep 21 '13 at 13:40
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HINT:

$$2x^3+x^2-3=2(x^3-1)+(x^2-1)=2\{(x-1)(x^2+x+1)\}+(x-1)(x+1)$$

$$=(x-1)\{2(x^2+x+1)+x+1\}=(x-1)(2x^2+3x+3)$$

Alternatively, using Remainder Theorem, $(x-1)|(2x^3+x^2-3)$

So, by actual division $2x^3+x^2-3=(x-1)(2x^2+3x+3)$

Do you know how to determine the nature of a Quadratic Equation?

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You could use the Rational Root Theorem to narrow down your possibilities of real solutions. Have you learned this?

Also, you could then factor out the factor with the real root by synthetic division or long division, and find the (imaginary) roots of the remaining factor using the quadratic formula.

I hope this helps!

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you could also guess one root and later divide the polynomial by the expression $x-x_0$. Guessing is based upon the fact that any rational root $\frac{p}{q}$ ($GCD(p,q) = 1$)satisfies the property: $p$ divides the coefficient at $x^0$ and $q$ divides the coefficient at $x^n$ where $n$ is the degree of the polynomial. In this case the guess $x = 1$ is straightforward... Also you could note that any polynomial of odd degree has at least one real root because complex roots come in pairs with their conjugates.

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