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Any sensible numerical method will lead to a solution $-16.3(2)$.
In particular, an interval-halving method is efficient and simple to implement. You can make it a little quicker by choosing a sensible starting point: note that any solution would have to be negative, since all of the coefficients are positive; another moment of inspection will also show that the solution must lie between $-10$ and $-100$, giving you a sensible starting point for a computation.
To determine whether there are any other solutions, note that the expression is a cubic and will therefore have either $1$ or $3$ real solutions.
To explore the properties of the cubic $y=2x^3+34x^2+567x+8901$, look at the turning points. Differentiating, we find that
$$
\frac{dy}{dx} = 6x^2+68x+567
$$
The discriminiant of this quadratic is $68^2-4\times 6 \times 567 = -8984$. Since this is negative, there are no turning points and the cubic consequently only has $1$ real solution.
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Age 16 to 18
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Any sensible numerical method will lead to a solution $-16.3(2)$.
In particular, an interval-halving method is efficient and simple to implement. You can make it a little quicker by choosing a sensible starting point: note that any solution would have to be negative, since all of the coefficients are positive; another moment of inspection will also show that the solution must lie between $-10$ and $-100$, giving you a sensible starting point for a computation.
To determine whether there are any other solutions, note that the expression is a cubic and will therefore have either $1$ or $3$ real solutions.
To explore the properties of the cubic $y=2x^3+34x^2+567x+8901$, look at the turning points. Differentiating, we find that
$$
\frac{dy}{dx} = 6x^2+68x+567
$$
The discriminiant of this quadratic is $68^2-4\times 6 \times 567 = -8984$. Since this is negative, there are no turning points and the cubic consequently only has $1$ real solution.
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