Find a polynomial with integer coefficients that satisfies the given conditions Q has degree 3 and zeros 3, 3i, and _3i. Q has... (answered by CubeyThePenguin). Answered by ishagarg. So now we have all three zeros: 0, i and -i.
The standard form for complex numbers is: a + bi. We will need all three to get an answer. This is our polynomial right. Another property of polynomials with real coefficients is that if a zero is complex, then that zero's complex conjugate will also be a zero. Q has degree 3 and zeros 4, 4i, and −4i. Q(X)... (answered by edjones). So in the lower case we can write here x, square minus i square. To create our polynomial we will use this form: Where "a" can be any non-zero real number we choose and the z's are our three zeros. In standard form this would be: 0 + i. There are two reasons for this: So we will multiply the last two factors first, using the pattern: - The multiplication is easy because you can use the pattern to do it quickly. The factor form of polynomial. These are the possible roots of the polynomial function. But we were only given two zeros. Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
For given degrees, 3 first root is x is equal to 0. Find every combination of. So it complex conjugate: 0 - i (or just -i). Q has... (answered by tommyt3rd). Find a polynomial with integer coefficients and a leading coefficient of one that... (answered by edjones). Total zeroes of the polynomial are 4, i. e., 3-3i, 3_3i, 2, 2. Since integers are real numbers, our polynomial Q will have 3 zeros since its degree is 3. By clicking Sign up you accept Numerade's Terms of Service and Privacy Policy. Explore over 16 million step-by-step answers from our librarySubscribe to view answer. It is given that the polynomial R has degree 4 and zeros 3 − 3i and 2.
Fusce dui lecuoe vfacilisis. X-0)*(x-i)*(x+i) = 0. This is why the problem says "Find a polynomial... " instead of "Find the polynomial... ". That is plus 1 right here, given function that is x, cubed plus x. Since we want Q to have integer coefficients then we should choose a non-zero integer for "a". Q has... (answered by Boreal, Edwin McCravy). Let a=1, So, the required polynomial is. We have x minus 0, so we can write simply x and this x minus i x, plus i that is as it is now. Pellentesque dapibus efficitu. In this problem you have been given a complex zero: i.
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