So a question give that P(x) is an odd polynomial of degree three. So now I know that $P(x) = ax^3 + bx^2 +cx +d $. And $P(-x) = -P(x) $since it is odd. However, the solutions say that given this informations, it can actually be written simply in the form $ax^3 +bx$. Why is this so?
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1do you mean $bx^2$ or $b^2$ when you define $P(x)$? – C Squared Nov 10 '20 at 08:49
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$P(-x)$ and $-P(x)$ are polynomials. If they are equal they must have the same coefficients. – Kavi Rama Murthy Nov 10 '20 at 08:51
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The equation $P(-x)=-P(x)$ says that the polynomial $P(x)$ is pointsymmetric to the origin.
If you plug in, $P(-x)= -ax^3+bx^2-cx+d$ and $-P(x)=-ax^3-bx^2-cx-d$. Then the equation gives $P(x) = ax^3+cx$.
Wuestenfux
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Also what did you do? Did you just equate the parts which are equal and discard the parts that are not? – user71207 Nov 10 '20 at 09:01
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For odd polynomial $P(x)$, we have $P(x)+P(-x)=0$ for all real values of $x$, so if $P(x)=ax^3+bx^2+cx+d$, then we get $$ax^3+bx^2+cx+d-ax^3+bx^2-cx+d=0 \implies 2bx^2+2d=0=0x^2+0,$$ for all real values of $x$. This implies $b=0=d$. So finally we have $P(x)=ax^3+cx$
Z Ahmed
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Thanks, I see. The equation P(x) + P(-x) = 0 is really insightful, I've never thought of it that way before – user71207 Nov 10 '20 at 08:57
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Ahhhhhhhhh I think I know. It is because only the distinct zero polynomial can be P(x) = 0 – user71207 Nov 10 '20 at 09:03
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When $bx^2+d=0 \implies x^2=-d/b$, bu this is called root(s). But if $bx^2+d=0$ for all real values of $x$ the $bx^2+d=0x^2+0$. Namely, coefficient of each powers of $x$ have to be zero. For instance if $px^2+qx+r=0, \forall x\in R$, this means $p=q=r=0$. But if it is an equation then $x={-q\pm\sqrt{q^2-4pr}}{2P}.$. I have edited my answer slightly for this point for you. – Z Ahmed Nov 10 '20 at 09:10