Plz help, will give medal Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 5, -3, and -1 + 2
god damnit I keep messing up the problem sorry hisham
??
5, -3, and -1 + 2i
1) f(x) = x4 - 4x3 + 10x2 + 20x + 75 2) f(x) = x4 - 14x2 - 40x - 75 3) f(x) = x4 - 4x3 - 10x2 - 20x - 75 4) f(x) = x4 + 10x2 - 40x - 75
ahh now it makes a lot of sense
can you look it again
1) f(x) = x^4 - 4x^3 + 10x^2 + 20x + 75 2) f(x) = x^4 - 14x^2 - 40x - 75 3) f(x) = x^4 - 4x^3 - 10x^2 - 20x - 75 4) f(x) = x^4 + 10x^2 - 40x - 75
i mean are you sure there is no -2i?
yes
it does not make sense to have only one i term!
im looking at it now, thats definitely what it says
It says that the function must have those listed, but it doesnt mean it cant have zeroes that arent listed there as well
ohhhh ok ok
x^5 - x^4 -13x^3 - 19x^2 -68x -60
I have to choose one of the 4 options
you also must have the polynomial raised the fifth power.
so which of the 4 options is it?
none
i need to choose one of those haha its multiple choice
to give you an idea of what's happening, you have 5 zeroes listed thus you must have the x raised to the power of 5
This is a polynomial of degree 4. To find zeros for polynomials of degree 3 or higher we use Rational Root Test. The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction pq, where p is a factor of the trailing constant and q is a factor of the leading coefficient. The factor of the leading coefficient (1) is 1 .The factors of the constant term (-75) are 1 3 5 15 25 75 . Then the Rational Roots Tests yields the following possible solutions: ±11, ±31, ±51, ±151, ±251, ±751 Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers. If we plug these values into the polynomial P(x), we obtain P(−3)=0. To find remaining zeros we use Factor Theorem. This theorem states that if pq is root of the polynomial then this polynomial can be divided with qx−p. In this example: Divide P(x) with x+3 x4−14x2−40x−75x+3=x3−3x2−5x−25 Polynomial x3−3x2−5x−25 can be used to find the remaining roots. Use the same procedure to find roots of x3−3x2−5x−25 When you get second degree polynomial use step-by-step quadratic equation solver (opens new window) to find two remaining roots.
found it online, thanks though you still get a medal for helping me with all these questions!
you're welcome. :)
but how does that help?
The roots of the polynomial x4−14x2−40x−75 are: x1=−3 x2=5 x3=−1+2i x4=−1−2i
-_-
oh my god!
I was solving for -1 and 2i instead of -1+2i
ohhh XD
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