Solving Non-Quadratic Polynomials

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Math › Solving Non-Quadratic Polynomials

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Consider the equation $14x^{4} +Ax^{3}+Bx^{2}+Cx+6 = 0$ .

According to the Rational Zeroes Theorem, if are all integers, then, regardless of their values, which of the following cannot be a solution to the equation?

$x=\frac{10}{3}$

$x=\frac{1}{2}$

$x = \frac{6}{7}$

$x= \frac{3}{14}$

Explanation

By the Rational Zeroes Theorem, any rational solution must be a factor of the constant, 6, divided by the factor of the leading coefficient, 14.

Four of the answer choices have this characteristic:

$3 = 3 \div 1$

$\frac{1}{2} = 1\div 2$

$\frac{6}{7} = 6 \div 7$

$\frac{3}{14} =3 \div14$

$\frac{10}{3}$ is in lowest terms, and 3 is not a factor of 14. It is therefore the correct answer.

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