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Miracles
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can a non-factorable function has rational real roots?
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Yes, for example take [itex]f(x):= x+a[/itex] where [itex]a \in \mathbb{Q}[/itex].Miracles said:can a non-factorable function has rational real roots?
Yes, a non-factorable function can have rational real roots. A rational real root is a value that, when plugged into the function, makes the function equal to zero. This can happen even if the function cannot be factored into simpler expressions.
You can determine if a non-factorable function has rational real roots by graphing the function or by using the rational root theorem. The rational root theorem states that if a polynomial function has rational roots, they will be in the form of p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. By checking all possible combinations of p and q, you can determine if the function has any rational roots.
Yes, a non-factorable function can have irrational real roots. An irrational real root is a value that cannot be expressed as a fraction and, when plugged into the function, makes the function equal to zero. These roots may not be obvious when looking at the function, but they can be found by using methods such as the quadratic formula or graphing the function.
No, rational real roots are not the only type of real roots a non-factorable function can have. In addition to rational real roots and irrational real roots, a non-factorable function can also have complex roots. These are values that involve the imaginary number i and cannot be expressed as real numbers. Complex roots always come in pairs, so if a non-factorable function has one complex root, it will also have another.
Yes, a non-factorable function can have no real roots. This means that there are no values that can be plugged into the function to make it equal to zero. Graphically, this would mean that the function does not cross the x-axis. Non-factorable functions with no real roots often have complex roots or no roots at all.