Find a factorable expression.

Algebra Level 3

Which of the following quadratic polynomials can be expressed as a product of two linear factors with rational coefficients?

4 x 2 + 6 x + 2 4x^{2} + 6x + 2 . 5 x 2 7 x 2 5x^{2} - 7x - 2 . x 2 3 x^{2} - 3 . 2 x 2 x + 3 2x^{2} - x + 3 . 3 x 2 4 x 1 3x^{2} -4x - 1 .

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1 solution

. .
Feb 18, 2021

Only 4 x 2 + 6 x + 2 4x ^ { 2 } + 6x + 2 is factorable because 4 x 2 + 6 x + 2 = ( 2 x + 1 ) ( 2 x + 2 ) = 2 ( x + 1 ) ( 2 x + 1 ) 4x ^ { 2 } + 6x + 2 = ( 2x + 1 ) ( 2x + 2 ) = 2 ( x + 1 ) ( 2x + 1 ) .

x 2 3 = ( x 3 ) ( x + 3 ) x^2 - 3 = ( x - \sqrt3)(x+ \sqrt3 )

Jason Gomez - 3 months, 3 weeks ago

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Factor them in the range of Rational numbers.

. . - 3 months, 3 weeks ago

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You could mention that in the question

Jason Gomez - 3 months, 3 weeks ago

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@Jason Gomez Formally, factoring means dividing the expressions with 2 or more than the factors, and until the rational number range.

. . - 3 months, 3 weeks ago

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@. . Technically, "factors" is not the same here as in number theory, which involves integers. See the examples here . While "factorization" involves factors in integer representations, that is not the rigorous definition. There exist some polynomials with recognizable patterns . Using one of the choices you listed, the simpler case is x 2 3 = ( x 3 ) ( x + 3 ) x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3}) . It is not factorizable in integer forms, but it is factorizable in real value forms. While the problem looks trivial, it isn't as all of the polynomials are factorizable in complex forms. That can easily be expressed by finding two roots (as that is the most possible number of roots of polynomials of degree 2 2 ) in the complex plane.

Also read this , which explores solutions of quadratic equations in a more complex perspective.

Michael Huang - 3 months, 3 weeks ago

But did you got correct?

. . - 3 months, 3 weeks ago

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Me? Yes I did but this is for others who come upon this question

Jason Gomez - 3 months, 3 weeks ago

Even 3 x 2 4 x 1 , 5 x 2 7 x 2 3x^2-4x-1,5x^2-7x-2 are factorable

Jason Gomez - 3 months, 3 weeks ago

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They are unfactorable in the range of rational numbers.

. . - 3 months, 3 weeks ago

All of these quadratics are factorable in the real numbers except for Choice D. You should consider restating the problem as which expression is irreducible (non-factorable) for real x.

tom engelsman - 3 months, 3 weeks ago

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I cannot change the answer choices after I posted a question.

. . - 3 months, 3 weeks ago

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U can change the question tho

Jason Gomez - 3 months, 3 weeks ago

And did you got correct?

. . - 3 months, 3 weeks ago

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