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anemone
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Prove that there are no integers $a,\,b,\,c$ and $d$ such that the polynomial $ax^3+bx^2+cx+d$ equals 1 at $x=19$ and 2 at $x=62$.
because this is a challenge question you are required to answer it fully . this is not a question for helpCountry Boy said:Well, to start with, since $ax^3+ bx^2+ cx+ d$ is 1 when x 19, $a(19)^3+ b(19)^2+ 19x+ d= 6859a+ 361b+ 19c+ d= 1$, And since it is 2 when x= 62, $a(62)^3+ b(62)^2+ a(62)+ d= 238328a+ 3844b+ 62c+ d= 2$,
Subtracting the first from the second, 231469a+ 3461b+ 53c= 1. Since a, b, and c are integers that is a linear Diophantine equation.
No, this equation cannot be solved for any integer value of x. It only has solutions for specific values of x, such as 19 and 62.
The number 1 is the constant term in this equation and represents the desired outcome or solution.
This equation has two solutions, one for x=19 and one for x=62.
The degree of the polynomial, in this case being 3, indicates the highest power of x in the equation. For a polynomial of degree n, there can be at most n solutions.
Yes, this equation can also be solved using factoring, the rational root theorem, or other algebraic methods. However, substitution is the most straightforward approach for solving this particular equation.