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**6.7 – Using the Fundamental Theorem of Algebra**

Objectives: Use the fundamental theorem of algebra to determine the number of zeros of a polynomial function. Write a polynomial function given the zeros of the function.

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**Review: Find all the zeros: f(x) = x3 + x2 – 2x – 2 Answer: , , -1**

A polynomial to the nth degree will have n zeros. f(x) = x3 – 6x2 – 15 x = (x + 4)(x – 5)(x – 5) the zeros are: -4, 5, 5 5 is a repeated solution

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**Review: Find zeros of functions/solutions to polynomial equations**

1. Factor the polynomial 2. Rational Zero Theorem (p/q), then factor using synthetic division 3. Set unsolved factors to zero and solve f(x) = x4 + 4x3 – 6x2 – 36x + 27 f(x) = x4 – 5x2 – 36

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**1. Use the fundamental theorem to determine the number of roots**

The Fundamental Theorem of Algebra If f(x) is a polynomial of degree n where n > 0, then the equation f(x) = 0 has at least one root in the set of complex numbers. That means if there is a variable, there is at least one solution in the complex numbers, a + bi.

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**1. Use the fundamental theorem to determine the number of roots**

We know: the degree of the equation tells you the number of solutions imaginary solutions come in pairs A new thing to know: if an imaginary number is a zero, then its conjugate is also a zero so, if 4 + 3i is solution, then 4 – 3i is also a solution

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**Using a Graphing Calculator to Find the Real Zeros**

Under “Y =” type in the equation. Go to [2nd]; “CALC”; “2: zero” Left bound: you need to place the cursor to the left of the intersection and press enter. Right bound: you need to place the cursor to the right of the intersection and press enter; and enter again. Type in the equation y = x3 + 3x2 + 16x + 48.

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**Example: find all zeros of x3 + 3x2 + 16x + 48 = 0**

(should be 3 solutions total) CT = ±1 ±2 ±3 ±4 ±6 ±8 ±12 ±16 ±24 ±48 LC ±1 Graph and you’ll see -3 is the only one on the graph you can see so synthetic divide with -3 x = 0 x2 = -16 x = ±√-16 = ±4i The three zeros are -3, 4i, -4i

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**Writing Polynomial Functions:**

Now write a polynomial function of least degree that has real coefficients, a leading coefficient of 1 and 1, -2 + i, -2 – i as zeros. f(x) = (x – 1)(x – (-2 + i))(x – (-2 – i)) f(x) = (x – 1)(x + 2 – i)(x i) f(x) = (x – 1){(x + 2) – i} {(x + 2) +i} f(x) = (x – 1){(x + 2)2 – i2} FOIL f(x) =(x – 1)(x2 + 4x + 4 –(-1)) Take care of i2 f(x) = (x – 1)(x2 + 4x ) f(x) = (x – 1)(x2 + 4x + 5) Multiply f(x) = x3 + 4x2 + 5x – x2 – 4x – 5 f(x) = x3 + 3x2 + x – 5

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**Writing Polynomial Functions:**

Now write a polynomial function of least degree that has real coefficients, a leading coefficient of 1 and 4, 4, 2 + i as zeros. Note: 2+i means 2 – i is also a zero f(x) = (x – 4)(x – 4)(x – (2 + i))(x – (2 – i)) f(x) = (x – 4)(x – 4)(x – 2 – i)(x – 2 + i) f(x) = (x2 – 8x + 16)((x – 2) – i)((x – 2) + i) f(x) = (x2 – 8x +16)((x – 2)2 – i2) f(x) = (x2 – 8x +16)(x2 – 4x + 4 –(-1)) f(x) = (x2 – 8x +16)(x2 – 4x + 5) f(x) = x4 – 4x3 + 5x2 – 8x3 + 32x2 – 40x + 16x2 – 64x + 80 f(x) = x4 – 12x3 + 53x2 – 104x + 80

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Assignment: pgs #15, 16, 21, 38, 39, 41, 42, 47

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