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Let f(x) = x^3 + 9x^2 + 23x +15
\n" ); document.write( "It's important to remember that a third order (or any odd order) polynomial must have at least one real root. That is, the function must \"cross\" the x-axis at least once. Think about it. When x is negative infinity x^3 is negative infinity. And when x is plus infinity, x^3 is plus infinity. As x goes from negative infinity to plus infinity, f(x) also goes from negative infinity to plus infinity. How can f(x) do that without crossing the x-axis? It can't. Therefore f(x) has at least one real root. Comment: A third order can cross the x-axis as many as three times - yielding three real roots. And in all cases there are three roots.
\n" ); document.write( "This fact being said, our first order o business is to find this real root. What I do is evaluate f(x) for different values of x until the function goes from positive to negative or visa versa. Then we know it crossed zero at some value of x in between the two values I tried.
\n" ); document.write( "I always start at x=0, in your case
\n" ); document.write( "f(0) = 0^3 + 9*0^2 + 23*0 + 15 = +15 > 0
\n" ); document.write( "Now try x = 1
\n" ); document.write( "f(1) = 1 + 9 + 23 + 15 = 48 > 0
\n" ); document.write( "Actually, there is no need to try positive values of x, because all terms in f(x) are positive. So as x goes more positive so does f(x). In order for f(x) = 0 some term(s) must be negative.
\n" ); document.write( "Let's try x = -1, Then
\n" ); document.write( "f(-1) = -1 + 9 -23 +15 = 0. Voila, we found the real root x = -1.
\n" ); document.write( "Our first factor is
\n" ); document.write( "(1) (x+1)
\n" ); document.write( "Note that the sign of the root is opposite to that of the number in the factor.
\n" ); document.write( "Now how do we find the other two factor? We simply divide f(x) by (x+1)
\n" ); document.write( "(x^3+9x^2+23x+15)/(x+1) = (ax^2 +bx +c)
\n" ); document.write( "Can you get find the values a and c?
\n" ); document.write( "Try cross multiplying
\n" ); document.write( "(x+1)(ax^2 +bx +c) = (x^3 +9x^2 +23x +15), then equate coefficients of like powers of x. Multiplying the left side yilds
\n" ); document.write( "ax^3 +bx^2 +ax^2 +cx +bx +c = 1x^3 +9x^2 +23x +15> Then
\n" ); document.write( "(1) a = 1
\n" ); document.write( "(2) b + a = 9
\n" ); document.write( "(3) c + b = 23
\n" ); document.write( "(4) c = 15
\n" ); document.write( "Put a and c into (2) and (3)
\n" ); document.write( "(2) b + 1 = 9 or
\n" ); document.write( "(5) b = 8
\n" ); document.write( "(3) 15 + b = 23 or
\n" ); document.write( "(5) b = 8
\n" ); document.write( "Our quadratic is
\n" ); document.write( "(6) x^2 +8x +15, which further factors to
\n" ); document.write( "(7) (x+5)(x+3)
\n" ); document.write( "Your trinomial factors to
\n" ); document.write( "(x+1)(x+3)(x+5)
\n" ); document.write( "Check this by FOIL
\n" ); document.write( "(x+1)(x^2 +8x +15) =
\n" ); document.write( "x^3 +8x^2 +15x +x^2 +8x +15 =
\n" ); document.write( "f(x) = x^3 +9x^2 +23x +15 \n" ); document.write( "