Question
Check whether or not is a field.
Answer :
Word Count : 651
We are asked to check whether ( \mathbb{Q}[x] / \langle 8x^3 + 6x^2 - 9x + 24 \rangle ) is a field. A quotient ring ( \mathbb{Q}[x]/\langle f(x) \rangle ) is a field if and only if the polynomial ( f(x) ) is irreducible over ( \mathbb{Q} ). Here, ( f(x) = 8x^3 + 6x^2 - 9x + 24 ). Step 1: Try Rational Root Theorem The Rational Root Theorem states that any rational root ( \frac{p}{q} ) (in lowest terms) satisfies ( p \mid 24 ) and ( q \mid 8 ). Divisors of 24: ( \pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24 ) Divisors of 8: ( \pm1, \pm2, \pm4, \pm8 ) So possible rational roots are ( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}, \dots ) etc. Step 2: Test simple integer roots * ( x = 1 ): ( 8 _________ ___ __________ ____ _________ _______ __________ __________ ________ ___ ________.
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We are asked to check whether ( \mathbb{Q}[x] / \langle 8x^3 + 6x^2 - 9x + 24 \rangle ) is a field. A quotient ring ( \mathbb{Q}[x]/\langle f(x) \rangle ) is a field if and only if the polynomial ( f(x) ) is irreducible over ( \mathbb{Q} ). Here, ( f(x) = 8x^3 + 6x^2 - 9x + 24 ). Step 1: Try Rational Root Theorem The Rational Root Theorem states that any rational root ( \frac{p}{q} ) (in lowest terms) satisfies ( p \mid 24 ) and ( q \mid 8 ). Divisors of 24: ( \pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24 ) Divisors of 8: ( \pm1, \pm2, \pm4, \pm8 ) So possible rational roots are ( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}, \dots ) etc. Step 2: Test simple integer roots * ( x = 1 ): ( 8 _________ ___ __________ ____ _________ _______ __________ __________ ________ ___ ________.
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