According to the rational root theorem, what are all the potential rational roots of f(x)= 5x^3-7x+11

1 answer:

8 0

The rational root theorem states that the rational roots of a polynomial can only be in the form p/q, where p divides the constant term, and q divides the leading term.

In your case, both the leading term 5 and the constant term 11 are primes, so their only divisors are 1 and themselves.

So, the only feasible solutions are

$\pm\dfrac{11}{5},\quad\pm 11,\quad \pm\dfrac{1}{5}, \pm 1$

For the record, in this case, none of the feasible solutions are actually a root of the polynomial.

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Rationalize the denominator of 4/14 root 2?

Katarina [22]

First of all, you can simplify the 4 at the numerator and the 14 at the denominator (they're both multiple of 2):

$\dfrac{4}{14\sqrt{2}} = \dfrac{2\cdot 2}{2\cdot 7\cdot\sqrt{2}} = \dfrac{2}{7\sqrt{2}}$

Now, rationalize a denominator means that you have to get rid of the square root, in order to have an integer denominator.

To do so, remember that you can always multiply any number by 1 without changing its value, and you can always think of 1 as a fraction where numerator and denominator are equal:

$\dfrac{2}{7\sqrt{2}} = \dfrac{2}{7\sqrt{2}}\cdot 1 = \dfrac{2}{7\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{2\sqrt{2}}{7\cdot 2} = \dfrac{\sqrt{2}}{7}$