Use the rational zero theorem to create a list of all possible rational zeroes of the function f(x) = 14x^7 - 4x^2 + 2

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lisboa lisboa · Answer 1 · 2018-10-29T23:48:51+01:00

[tex]f(x) = 14x^7 - 4x^2 + 2[/tex]

Use the rational zero theorem

In rational zero theorem, the rational zeros of the form +-p/q

where p is the factors of constant

and q is the factors of leading coefficient

[tex] f(x) = 14x^7 - 4x^2 + 2[/tex]

In our f(x), constant is 2 and leading coefficient is 14

Factors of 2 are 1, 2

Factors of 14 are 1,2, 7, 14

Rational zeros of the form +-p/q are

[tex]+-\frac{1,2}{1,2,7,14}[/tex]

Now we separate the factors

[tex]+-\frac{1}{1}, +-\frac{1}{2}, +-\frac{1}{7}, +-\frac{1}{14},+-\frac{2}{1}, +-\frac{2}{2}, +-\frac{2}{7}, +-\frac{2}{14} [/tex]

[tex]+-1, +-\frac{1}{2}, +-\frac{1}{7}, +-\frac{1}{14},+-2, +-1 , +-\frac{2}{7}, +-\frac{1}{2} [/tex]

We ignore the zeros that are repeating

[tex]+-1, +-2, +-\frac{1}{2}, +-\frac{1}{7}, +-\frac{1}{14}, +-\frac{2}{7} [/tex]

Option A is correct

slicergiza slicergiza · Answer 2 · 2019-10-24T08:17:28+02:00

Answer:

[tex]\pm 1, \pm 2 \pm \frac{1}{2},\pm \frac{1}{7},\pm \frac{1}{14}, \pm \frac{2}{7}[/tex]

Step-by-step explanation:

By the rational root theorem or rational zero theorem,

The possible;e solutions of a polynomial function is,

[tex]\pm(\frac{\text{factors of the constant term}}{\text{Factors of the leading coefficient}})[/tex]

Here, the given function,

[tex]f(x) = 14x^7 - 4x^2 + 2[/tex]

Constant term = 2 and Leading coefficient = 14,

Factors of 2 = 1, 2,

Factors of 14 = 1, 2, 7, 14

Hence, the possible roots of the function,

[tex]\pm(\frac{1, 2}{1, 2, 7, 14})[/tex]

[tex]=\pm 1, \pm 2, \pm \frac{1}{2},\pm \frac{1}{7},\pm \frac{1}{14},\pm \frac{2}{7}[/tex]