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Thread: Quad Factoring Problem

  1. #1
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    Post Quad Factoring Problem

    $\displaystyle x^2+x+1=0$
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Pre-Calculus View Post
    $\displaystyle x^2+x+1=0$
    $\displaystyle x=\frac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}$

    It ain't looking good...
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  3. #3
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    What do you mean?
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Pre-Calculus View Post
    What do you mean?
    I mean that this particular equation has no real zeros.

    IE $\displaystyle x^2+x+1\neq0$ for all x.

    The discriminant in the quadratic formula conveys this.
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  5. #5
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    Quote Originally Posted by Pre-Calculus View Post
    $\displaystyle x^2+x+1=0$
    Normally, "factoring" means "factor into terms having integer coefficients" and, as VonNemo19 said, that cannot be done here.

    But if you really want to factor- and aren't concerned about the factors having integer, or even real, coefficients, you can "complete the square":

    $\displaystyle x^2+ x+ 1= x^2+ x+ \frac{1}{4}- \frac{1}{4}+ 1= 0$
    (Divide the coefficient of x (1) by to (1/2) and square (1/4). Add and subtract that.)

    $\displaystyle x^2+ x+ 1= (x+ \frac{1}{2})^2+ \frac{3}{4}$
    and think of this as a "difference of squares":

    $\displaystyle (x+ \frac{1}{2})^2- (-\frac{3}{4})= (x+\frac{1}{2})^2- (i \frac{\sqrt{3}}{2})^2$

    Which can be factored as "sum and difference": $\displaystyle a^2- b^2= (a-b)(a+b)$.

    $\displaystyle x^2+ x+ 1= (x+ \frac{1}{2}- i\frac{\sqrt{3}}{2})(x+\frac{1}{2}+ i\frac{\sqrt{3}}{2})$
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