How do you factor $x^{2} - 3 x + 15$ $x^2-3x+15$ ?

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How do you factor $x^{2} - 3 x + 15$ $x^2-3x+15$ ?

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Answer 1 · 2015-08-08T12:27:46

Examine to discriminant to find that this has no factors with Real coefficients. Use the quadratic formula to find:

$x^{2} - 3 x + 15 = (x - \frac{3}{2} - i \frac{\sqrt{51}}{2}) (x - \frac{3}{2} + i \frac{\sqrt{51}}{2})$ $x^2-3x+15 = (x-3/2-isqrt(51)/2)(x-3/2+isqrt(51)/2)$

Explanation:

$x^{2} - 3 x + 15$ $x^2-3x+15$ is of the form $a x^{2} + b x + c$ $ax^2+bx+c$ with $a = 1$ $a=1$ , $b = - 3$ $b=-3$ and $c = 15$ $c=15$ .

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = (-3)^2-(4xx1xx15) = 9 - 60 = -51$

Since $Delta < 0$ , $x^2-3x+15 = 0$ has no Real roots and $x^2-3x+15$ has no factors with Real coefficients.

The complete conjugate roots of $x^2-3x+15 = 0$ are given by the quadratic formula:

$x = (-b+-sqrt(Delta))/(2a) = (3+-isqrt(51))/2$

So $x^2-3x+15 = (x-3/2-isqrt(51)/2)(x-3/2+isqrt(51)/2)$

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How do you factor $x^{2} - 3 x + 15$ $x^2-3x+15$ ?

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