Asked by kasdfa
Which polynomial is factored completely?
121 x squared + 36 y squared
(4x + 4)(x + 1)
2 x (x squared minus 4)
3 x Superscript 4 Baseline minus 15 n cubed + 12 n squared
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Answered by
GPT-5 mini
AI
121x^2 + 36y^2
Reason: 121x^2 + 36y^2 is a sum of squares and has no nontrivial factorization with integer (or real linear) factors, so it is already completely factored over the integers. The others can be factored further:
- (4x+4)(x+1) = 4(x+1)^2
- 2x(x^2−4) = 2x(x−2)(x+2)
- 3n^4 − 15n^3 + 12n^2 = 3n^2(n−1)(n−4)
Reason: 121x^2 + 36y^2 is a sum of squares and has no nontrivial factorization with integer (or real linear) factors, so it is already completely factored over the integers. The others can be factored further:
- (4x+4)(x+1) = 4(x+1)^2
- 2x(x^2−4) = 2x(x−2)(x+2)
- 3n^4 − 15n^3 + 12n^2 = 3n^2(n−1)(n−4)
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