Find the zeros of f(x) = x^3 +9x^2 + 23x + 15
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f(x) = x^3 +9x^2 + 23x + 15
To solve, we will subsitute with the 15 factors which are:
1, -1, 3, -3, 5, -5, 15, -15
let us subsitute with x= -5:
f(-5) = 5^3 + 9*5^2 + 23*5 + 15
= -125 + 225 - 529 + 15 = 0
The (x+5) is one of the factors:
==> f(x) = (x+ 5) * g(x)
Now we will divide f(x) by (x+5):
==> f(x) = (x+ 5) (x^2 + 4x + 3)
Now factor (x^2 + 4x + 3)
==> f(x) = (x+5)(x+3)(x+1)
Now we will determine the zeros:
==> x1= -5
==> x2= -3
==> x3= -1
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To find the zeros of f(x) = x^3 +9x^2 + 23x + 15.
We observe that if we put x= -1, then f(-1) = (-1)+9-23+15 = 0.
So x+1 is a facor of the f(x).
Therefore x^3+9x^2+23x+15 = (x+1))(x^2 +kx+15) , as this agrees with x^3 and constant terms.
Now equating x terms on both sides, 23x = 15x+kx.
kx = 23x-15x
kx = 8x.
k = 8.
Therefore x^2+8x+15 is a factor.
x^2+8x+15 = (x+5)(x+3).
Therefore f(x) = x^3+9x^2+15x+23 = (x+1)(x+3)(x+5).
Therefore x+1 = 0 , x+5 = 0 , x+3 = 0 gives the zeros of f(x)
Or x = -1, x= -5 , or x= -3 are the zeros of f(x).
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