SOLUTION: Factorise each of the following over C
a) f(z)=z^3+z^2+2z-4
b)g(z)= z^3+(2-i)z^2-z-2+i
i tried solving them as:
a) the factor of f(z) is (z-1), then do the long division and
Algebra.Com
Question 470838: Factorise each of the following over C
a) f(z)=z^3+z^2+2z-4
b)g(z)= z^3+(2-i)z^2-z-2+i
i tried solving them as:
a) the factor of f(z) is (z-1), then do the long division and the answer will be (z^2+2z+4) and so
f(z)=(z-1)(z^2+2z+4)
= (z-1)(z+1)^2 -1+4 <<< got stuck, how can i complete?
b)the factor of g(z) is (z+2+i), then do the long division and the answer will be z^2+1 and so
g(z)= (z+2-i)(z^2-1) <<<< how can i complete?
Thanks
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
On problem number 1, so far, so good.
\left(z^2\ +\ 2z\ + 4\right))
Just like you said. Next take the quadratic factor and set it equal to zero, then solve the quadratic equation using the quadratic formula:
^2\ -\ (4)(1)(4)}}{2(1)}\ =\ -1\ \pm\ i\sqrt{3})
.
Hence the factors of the quadratic trinomial are:
\right)\left(z\ -\ (-1\ -\ i\sqrt{3})\right))
Hence:
\left(z^2\ +\ 2z\ + 4\right)\ =\ (z\ -\ 1)\left(z\ +\ 1\ -\ i\sqrt{3}\right)\left(z\ +\ 1\ +\ i\sqrt{3}\right))
The other one is easier. Factor the quadratic factor as the difference of two squares.
John
My calculator said it, I believe it, that settles it
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