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FINDING A POLYNOMIAL PASSING THROUGH A POINT. Review: the Linear Factorization Theorem If where n > 1 and a n ≠ 0 then Where c 1, c 2, … c n are complex.

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Presentation on theme: "FINDING A POLYNOMIAL PASSING THROUGH A POINT. Review: the Linear Factorization Theorem If where n > 1 and a n ≠ 0 then Where c 1, c 2, … c n are complex."— Presentation transcript:

1 FINDING A POLYNOMIAL PASSING THROUGH A POINT

2 Review: the Linear Factorization Theorem If where n > 1 and a n ≠ 0 then Where c 1, c 2, … c n are complex numbers (possibly real and not necessarily distinct)  This theorem lets us generate polynomials with any zeroes by multiplying their corresponding factors.

3 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 (a n ) that results in the point we want.

4 The Technique 1. Determine all the zeroes you want your polynomial to have and what multiplicity each should have. 2. Generate a factor for each zero. 3. Multiply together all the factors. Multiply by each one a number of times equal to its multiplicity. 4. Plug in the point that you want the polynomial to pass through and determine the value of a n.

5 Example  Find a third degree polynomial with zeroes -1, 1, and 3 that passes through the point (2, -6).

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.

7 Solution: Step 2  Since our zeroes are -1, 1, and 3, our three factors are (x + 1), (x – 1), and (x – 3).

8 Solution: Step 3  Now, we multiply all the factors together and expand the product.  This gives us x 3 – 3x 2 – x + 3.  Our equation, in its current form, is f(x) = a n (x 3 – 3x 2 – x + 3).  Now we just need to find a n.  If we just wanted a polynomial with the given zeroes, we could choose any value of a n (usually 1), but we need our function to go through a particular point.

9 Solution: Step 4  To find a n, we plug in our point – (2, -6).  This gives us -6= a n (2 3 – 3*2 2 – 2 + 3). -6 = a n (8 – 12 – 2 + 3) -6 = -3a n a n = 2  Our final polynomial is f(x) = 2x 3 – 6x 2 – 2x + 6.

10 Graph We can check our answer by checking whether the graph of the function goes through the point we wanted. It does!


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