You can
put this solution on YOUR website!You know that roots result in zeros.
So a zero at x=-2 means (x+2) is a factor in the formula for the polynomial.
Likewise you are told you have an imaginary root at 7i. You know that imaginary roots come in pairs (7i , -7i)
So two more factors are (x-7i) and (x+7i)
Now you have three terms. The polynomial is the product of those three factors.
All you need to do now is simplify by multiplying it all out. I'll leave that to you, because I know you can do that. Ping me if you need more help
You can
put this solution on YOUR website!In general a polynomial function with zeros of
,
,
, ... ,
will have an equation of the form:
...
So if we know all the zeros of the polynomial then we will be able to find the polynomial. You are given two zeros but a polynomial of degree 3 should have 3 zeros. we need the third zero. The key is to understand when a zero is complex or imaginary, like 7i, then its conjugate, -7i, will also be a zero. So we now have our three zeros: -2, 7i and -7i.
With these zeros we can write the equation:
All that is left is to simplify. First we'll write the "minus a negatives" as additions of positives:
Now we'll multiply it out. It is easiest if we multiply the last two factors first because they fit the pattern
and because we want to get rid of the i's ASAP:
Since
and since
, we get:
And after the last multiplication we get:
You can
put this solution on YOUR website!Find a polynomial function of degree 3 with -2, and 7i as zeros.
----------------
If the polynomial function must have Real Number coefficients,
-7i must also be a zero.
-----------------------------
f(x) = (x-3)(x-7i)(x+7i)
f(x) = (x-3)(x^2+49)
f(x) + x^3-3x^2+49x-147
=============================
Cheers,
Stan H.
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