Find a third-degree polynomial with real coefficients and with zeros –3 and –4 + i.

Question

anonymous · Answer

one factor is $x+3$ for sure 
the hard part is finding a quadrtic with zeros $-4+i$ and$-4-i$ 
there are three ways do to it
one is hard
one is easy
one is real real easy
you pick

misty1212 · Answer

lol pick the hard way $\color\magenta\heartsuit$

anonymous · Answer

okay, lol.

anonymous · Answer

lets do the easy way first ]
work backwards starting with 
$$x=-4+i$$ then add 4 to both sides to get 
$$x+4=i$$ square both sides (carefully) to get $$(x+4)^2=-1$$ or $$x^2+8x+16=-1$$

anonymous · Answer

adding 1 gives you the quadratic as $$x^2+8x+17$$

anonymous · Answer

that is the easy wat

misty1212 · Answer

i guess that was easy...

anonymous · Answer

the real real easy way requires memorizing something
namely, that if $a+bi$ is a zero, the quadratic is $$x^2-2ax+(a^2+b^2)$$ in your case $a=-4,b=1$ so 
$$x^2-2(-4)x+(6^2+1^2)$$
$$x^2+8x+17$$

anonymous · Answer

typo there, but you get the idea

misty1212 · Answer

yeah $$x^2-2(-4)x+((-4)^2+1^2)$$

misty1212 · Answer

lot easier than trying $$(x-(-4+i))(x-(-4-i))$$

anonymous · Answer

final job is to multiply $$(x+3)(x^2+8x+17)$$

beginnersmind · Answer

$$(x-(-4+i))(x-(-4-i))$$
isn't hard either. It's in the form [(x+4)+i] * [(x+4)-i] which equals:

(x+4)^2 + 1