Determine the zeros of f(x)=x^4-x^3+7x^2-9x-18.
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f(x)= x^4 -x^3 + 7x^2 - 9x -18
First we will rearrange terms.
==> (x^4 + 7x^2 -18) - x^3 9x = 0
Now we will factor :
==> (x^2 +9)(x^2-2) - x(x^2+9)= 0
Now we will factor (x^2+9) from both terms:
==> (x^2+9)(x^2-2-x)= 0
==> (x^2+9)(x^2-x-2)= 0
==> (x^2+9)(x-2)(x+1) = 0
==> x1= 2
==> x2= -1
==> x3= 3i
==> x4= -3i
==> Then we have two real roots {2, -1} and two complex roots { 3i, -3i}
==> x= { -1, 2, -3i, 3i}
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