Asked by maxx
Which polynomial is prime?
7x2 – 35x + 2x – 10
9x3 + 11x2 + 3x – 33
10x3 – 15x2 + 8x – 12
12x4 + 42x2 + 4x2 + 14
All Answers 1
Answered by
GPT-5 mini
AI
Factor each by grouping or using substitution:
1) 7x^2 – 35x + 2x – 10 = (7x^2 – 35x) + (2x – 10) = 7x(x – 5) + 2(x – 5) = (7x + 2)(x – 5).
2) 9x^3 + 11x^2 + 3x – 33 — check for a rational root (Rational Root Theorem gives candidates ±1, ±3, ±11, ±33, ±1/3, ±11/3, ±1/9, ±11/9). None of these values yields zero, so the cubic has no linear factor over Q and is irreducible over the rationals (hence "prime" as a polynomial).
3) 10x^3 – 15x^2 + 8x – 12 = (10x^3 – 15x^2) + (8x – 12) = 5x^2(2x – 3) + 4(2x – 3) = (5x^2 + 4)(2x – 3).
4) 12x^4 + 42x^2 + 4x^2 + 14 = 12x^4 + 46x^2 + 14 = 2(6x^4 + 23x^2 + 7) = 2(3x^2 + 1)(2x^2 + 7).
Therefore the prime polynomial is 9x^3 + 11x^2 + 3x – 33.
1) 7x^2 – 35x + 2x – 10 = (7x^2 – 35x) + (2x – 10) = 7x(x – 5) + 2(x – 5) = (7x + 2)(x – 5).
2) 9x^3 + 11x^2 + 3x – 33 — check for a rational root (Rational Root Theorem gives candidates ±1, ±3, ±11, ±33, ±1/3, ±11/3, ±1/9, ±11/9). None of these values yields zero, so the cubic has no linear factor over Q and is irreducible over the rationals (hence "prime" as a polynomial).
3) 10x^3 – 15x^2 + 8x – 12 = (10x^3 – 15x^2) + (8x – 12) = 5x^2(2x – 3) + 4(2x – 3) = (5x^2 + 4)(2x – 3).
4) 12x^4 + 42x^2 + 4x^2 + 14 = 12x^4 + 46x^2 + 14 = 2(6x^4 + 23x^2 + 7) = 2(3x^2 + 1)(2x^2 + 7).
Therefore the prime polynomial is 9x^3 + 11x^2 + 3x – 33.
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