SOLUTION: factorise fully: x^3+9x^2+23x+15

Question 638426: factorise fully: x^3+9x^2+23x+15
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
Let f(x) = x^3 + 9x^2 + 23x +15
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.
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.
I always start at x=0, in your case
f(0) = 0^3 + 9*0^2 + 23*0 + 15 = +15 > 0
Now try x = 1
f(1) = 1 + 9 + 23 + 15 = 48 > 0
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.
Let's try x = -1, Then
f(-1) = -1 + 9 -23 +15 = 0. Voila, we found the real root x = -1.
Our first factor is
(1) (x+1)
Note that the sign of the root is opposite to that of the number in the factor.
Now how do we find the other two factor? We simply divide f(x) by (x+1)
(x^3+9x^2+23x+15)/(x+1) = (ax^2 +bx +c)
Can you get find the values a and c?
Try cross multiplying
(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
ax^3 +bx^2 +ax^2 +cx +bx +c = 1x^3 +9x^2 +23x +15> Then
(1) a = 1
(2) b + a = 9
(3) c + b = 23
(4) c = 15
Put a and c into (2) and (3)
(2) b + 1 = 9 or
(5) b = 8
(3) 15 + b = 23 or
(5) b = 8
Our quadratic is
(6) x^2 +8x +15, which further factors to
(7) (x+5)(x+3)
Your trinomial factors to
(x+1)(x+3)(x+5)
Check this by FOIL
(x+1)(x^2 +8x +15) =
x^3 +8x^2 +15x +x^2 +8x +15 =
f(x) = x^3 +9x^2 +23x +15
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