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Math Help - [SOLVED] How to show that the n factor is always positive?

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    [SOLVED] How to show that the n factor is always positive?

    Considering the polynomial, p(x) = 2x^4 + x^3 - 3x^2 - 2x -2 ,

    Show that the third factor of P(x) is always positive. The other two factors are x-\sqrt{2} and x+\sqrt{2}. Hence find the set values of x such that P(x)>0
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    Quote Originally Posted by mark1950 View Post
    Considering the polynomial, p(x) = 2x^4 + x^3 - 3x^2 - 2x -2 ,

    Show that the third factor of P(x) is always positive. The other two factors are x-\sqrt{2} and x+\sqrt{2}. Hence find the set values of x such that P(x)>0
    Show that the third factor of P(x) is always positive.

    divide p(x) by (x-\sqrt{2})(x+\sqrt{2}) = x^2-2

    analyze the quadratic quotient's discriminant

    ... find the set values of x such that P(x)>0

    solve the inequality x^2 - 2 > 0
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    Quote Originally Posted by mark1950 View Post
    Considering the polynomial, p(x) = 2x^4 + x^3 - 3x^2 - 2x -2 ,

    Show that the third factor of P(x) is always positive. The other two factors are x-\sqrt{2} and x+\sqrt{2}. Hence find the set values of x such that P(x)>0
    The product of the two given factors is x^2 - 2. Therefore p(x) = 2x^4 + x^3 - 3x^2 - 2x -2 = (x^2 - 2)(2x^2 ........ + 1) and it shouldn't be too hard to fill in the gap. Then note that the third factor is an irreducible quadratic ....
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    Oh thanks guys.
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    Quote Originally Posted by skeeter View Post
    Show that the third factor of P(x) is always positive.

    divide p(x) by (x-\sqrt{2})(x+\sqrt{2}) = x^2-2

    analyze the quadratic quotient's discriminant

    ... find the set values of x such that P(x)>0

    solve the inequality x^2 - 2 > 0
    I found out the third factor and did the discriminant thingy, b2 - 4ac and found that it's less than 0. This shows that it is always above the x-axis and therefore it is always positive. Is it correct or did I miss out something?
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    Quote Originally Posted by skeeter View Post
    ... find the set values of x such that P(x)>0

    solve the inequality x^2 - 2 > 0
    Um, if u don't mind me asking, why did you choose x^2 - 2 > 0 ? Can we choose other factors instead of that?

    Since they used the word, Hence, shouldn't we be using something from 2x^2 + 2x + 1?
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    Quote Originally Posted by mark1950 View Post
    I found out the third factor and did the discriminant thingy, b2 - 4ac and found that it's less than 0. This shows that it is always above the x-axis and therefore it is always positive. Is it correct or did I miss out something?
    If the leading term of an irreducible quadratic function q(x) is positive then q(x) > 0 for all x.

    If the leading term of an irreducible quadratic function q(x) is negative then q(x) < 0 for all x.

    Quote Originally Posted by mark1950 View Post
    Um, if u don't mind me asking, why did you choose x^2 - 2 > 0 ? Can we choose other factors instead of that?

    Since they used the word, Hence, shouldn't we be using something from 2x^2 + 2x + 1?
    The only thing that can make p(x) negative is the x^2 - 2 factor ....
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    Quote Originally Posted by mark1950 View Post
    Um, if u don't mind me asking, why did you choose x^2 - 2 > 0 ? Can we choose other factors instead of that?
    Since they used the word, Hence, shouldn't we be using something from 2x^2 + 2x + 1?
    By this point, you had already proved that 2x^2+ 2x+ 1, the "third factor" was always positive. That meant that whether the entire polynomial was positive or not depended entirely on the other two factors.
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