What is the completely factored form of f(x)=x3+5x2+4x−6?f(x)=(x−3)(x−(−1+i3√))(x−(−1−i3√))f(x)=(x+3)(x−(−1+i3√))(x−(−1−i3√))f(x)=(x+3)(x−(−1+3√))(x−(−1−3√))f(x)=(x−3)(x−(−1+3√))(x−(−1−3√))
Question
Answer:
[tex]f(x)=x^3+5x^2+4x-6[/tex][tex]f(-3)=(-3)^3+5(-3)^2+4(-3)-6=0\implies x+3\text{ is a factor of }f(x)[/tex]
Synthetic division yields
-3 | 1 5 4 -6
. | -3 -6 6
- - - - - - - - - - - - -
. | 1 2 -2 0
which translates to
[tex]\dfrac{x^3+5x^2+4x-6}{x+3}=x^2+2x-2[/tex]
with remainder 0. Now by the quadratic formula,
[tex]x^2+2x-2=0\implies x=\dfrac{-2\pm\sqrt{2^2-4(1)(-2)}}2=-1\pm\sqrt3[/tex]
and so
[tex]f(x)=x^3+5x^2+4x-6=(x+3)(x-(-1+\sqrt3))(x-(-1-\sqrt3))[/tex]
solved
general
10 months ago
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