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FINDING A POLYNOMIAL PASSING THROUGH A POINT Review: the Linear Factorization Theorem If where n > 1 and an ≠ 0 then Where c1, c2, … cn are complex numbers (possibly real and not necessarily distinct) This theorem lets us generate polynomials with any zeroes by multiplying their corresponding factors. Specific Points If we want to find a polynomial with given zeroes that passes through a point, we can use basically the same technique. We’ll still multiply together all the factors that correspond to the zeros of the polynomial, but we’ll also multiply by a constant term (an) that results in the point we want. The Technique 1. 2. 3. 4. Determine all the zeroes you want your polynomial to have and what multiplicity each should have. Generate a factor for each zero. Multiply together all the factors. Multiply by each one a number of times equal to its multiplicity. Plug in the point that you want the polynomial to pass through and determine the value of an. Example Find a third degree polynomial with zeroes -1, 1, and 3 that passes through the point (2, -6). Solution: Step 1 The first step is to determine all the zeroes and their multiplicities. With three zeros, we already have enough factors to form a third degree polynomial, so all of our zeros will have multiplicity 1. The zeroes, again, are -1, 1, and 3. Solution: Step 2 Since our zeroes are -1, 1, and 3, our three factors are (x + 1), (x – 1), and (x – 3). Solution: Step 3 Now, we multiply all the factors together and expand the product. This gives us x3 – 3x2 – x + 3. Our equation, in its current form, is f(x) = an(x3 – 3x2 – x + 3). Now we just need to find an. If we just wanted a polynomial with the given zeroes, we could choose any value of an (usually 1), but we need our function to go through a particular point. Solution: Step 4 To find an, we plug in our point – (2, -6). This gives us -6= an(23 – 3*22 – 2 + 3). -6 = an(8 – 12 – 2 + 3) -6 = -3an an = 2 Our final polynomial is f(x) = 2x3 – 6x2 – 2x + 6. Graph We can check our answer by checking whether the graph of the function goes through the point we wanted. It does!