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3 votes

3rd degree polynomial reconstruction

For the first condition you have $$ p(-1) p(1) \ge 0 $$ Assume $p(-1)p(1)>0$ so the $p(1)$ and $p(-1)$ are both positive or negative. Assume they are positive, then, by induction you can prove $$ p(...
Marco's user avatar
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2 votes
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Show that if $x=-1$ is a solution of $x^{3}-2bx^{2}-a^{2}x+b^{2}=0$, then $1-\sqrt{2}\le b\le1+\sqrt{2}$

You seem to have made a mistake during simplification and/or completing the square. Substituting $x = -1$ into the given equation, we get $$\begin{align*} -1 - 2b + a^2 + b^2 &= 0 \\[0.3cm] \...
Haris's user avatar
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2 votes

3rd degree polynomial reconstruction

If $p$ has only one real root the the root lies between two integers $k - 1$ and $k + 1$ where $p$ has opposite signs, violating condition 1. Therefore $p$ must have three real roots; call them $r_1 \...
David K's user avatar
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1 vote
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How to use the cubic formula.

In general, the real and imaginary parts of solutions to polynomial equations whose degree is $3$ or more can not be expressed as a finite sequence of arithmetic operations and $n$th root extractions (...
Michael Ejercito's user avatar
1 vote

Solve the cubic polynomial $a^3 - 10a + 5 = 0$

We can easily check that $a^3-10a+5=0$ has no rational solution. For polynomials with leading coefficient 1 and constant coefficient $5$, only $-5, -1, 1,$ and $5$ are possible rational solutions, due ...
Michael Ejercito's user avatar
1 vote

Solving a cubic polynomial equation.

Let us begin with $$ax^3+bx^2+cx+d=0$$ What we can do is substitute $x=y-\frac{b}{3a}$ $$a(y-\frac{b}{3a})^3+b(y-\frac{b}{3a})^2+c(y-\frac{b}{3a})+d=0$$ Relying on the binomial theorem, we can expand ...
Michael Ejercito's user avatar

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