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122 course ap precalculus segment 36a - finding polynomial equatio due …

Question

122
course
ap precalculus
segment
36a - finding polynomial equatio
due mon, oct 27, 2025
1a
default gallery

  1. ( y = x ^ { 4 } + 7 x ^ { 3 } + 18 x ^ { 2 } + 20 x + 8 )

a. is 5 a possible ration root of this question? how do you know?
text math set matrix
answer

Explanation:

Step1: Recall the Rational Root Theorem

The Rational Root Theorem states that if a polynomial \(P(x)=a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0\) has a rational root \(\frac{p}{q}\), then \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading - coefficient \(a_n\).

For the polynomial \(y=x^{4}+7x^{3}+18x^{2}+20x + 8\), \(a_n = 1\) (the coefficient of \(x^{4}\)) and \(a_0=8\).

Step2: Apply the Rational Root Theorem

If \(\frac{p}{q}\) is a rational root, then \(p\mid8\) (i.e., \(p=\pm1,\pm2,\pm4,\pm8\)) and \(q\mid1\) (i.e., \(q = \pm1\)). The possible rational roots are \(\frac{p}{q}=\pm1,\pm2,\pm4,\pm8\).

Since \(5\) is not a factor of \(8\) (when \(q = 1\)), \(5\) is not a possible rational root.

Answer:

No. By the Rational Root Theorem, for the polynomial \(y=x^{4}+7x^{3}+18x^{2}+20x + 8\) with leading - coefficient \(a_n = 1\) and constant term \(a_0 = 8\), the possible rational roots are of the form \(\frac{p}{q}\) where \(p\) is a factor of \(8\) (\(p=\pm1,\pm2,\pm4,\pm8\)) and \(q\) is a factor of \(1\) (\(q=\pm1\)). Since \(5\) is not a factor of \(8\), \(5\) is not a possible rational root.