Question

which of the following choices is not a possible rational root to the polynomial equation?
$f(x)=x^{4}-11x^{3}+41x^{2}-61x + 30$
1
2
3
4
use the rational root theorem to determine possible rational roots of the polynomial. or
to determine if a number is a root, try using synthetic division with that number to determine if the remainder is zero.

Explanation:

Step1: Apply Rational Root Theorem

The Rational Root Theorem states that if a polynomial \(a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0\) has integer - coefficients, then the possible rational roots are of the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading - coefficient \(a_n\). For the polynomial \(f(x)=x^4 - 11x^3 + 41x^2 - 61x + 30\), \(a_n = 1\) and \(a_0=30\). The factors of \(a_0 = 30\) are \(\pm1,\pm2,\pm3,\pm5,\pm6,\pm10,\pm15,\pm30\), and the factors of \(a_n = 1\) are \(\pm1\). So the possible rational roots are \(\pm1,\pm2,\pm3,\pm5,\pm6,\pm10,\pm15,\pm30\).

Step2: Test the given options

• For \(x = 1\):

\[

$$\begin{align*} f(1)&=(1)^4-11(1)^3 + 41(1)^2-61(1)+30\\ &=1-11 + 41-61 + 30\\ &=(1 + 41+30)-(11 + 61)\\ &=72 - 72\\ &=0 \end{align*}$$

\]

• For \(x = 2\):

\[

$$\begin{align*} f(2)&=(2)^4-11(2)^3 + 41(2)^2-61(2)+30\\ &=16-88 + 164-122 + 30\\ &=(16+164+30)-(88 + 122)\\ &=210 - 210\\ &=0 \end{align*}$$

\]

• For \(x = 3\):

\[

$$\begin{align*} f(3)&=(3)^4-11(3)^3 + 41(3)^2-61(3)+30\\ &=81-297+369 - 183+30\\ &=(81+369+30)-(297 + 183)\\ &=480 - 480\\ &=0 \end{align*}$$

\]

• For \(x = 4\):

\[

$$\begin{align*} f(4)&=(4)^4-11(4)^3 + 41(4)^2-61(4)+30\\ &=256-704+656-244 + 30\\ &=(256+656+30)-(704 + 244)\\ &=942 - 948\\ &=-6 eq0 \end{align*}$$

\]

Answer:

D. 4