1. EXAMPLE 3 Use zeros to write a polynomial function Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and 3 and 2 + 5 as zeros. SOLUTION Because the coefficients are rational and 2 + 5 is a zero, 2 – 5 must also be a zero by the irrational conjugates theorem. Use the three zeros and the factor theorem to write f (x) as a product of three factors.

2. EXAMPLE 3 Use zeros to write a polynomial function f (x) = (x – 3) [ x – (2 + √ 5 ) ] [ x – (2 – √ 5 ) ] Write f (x) in factored form. Regroup terms. = (x – 3) [ (x – 2) – √ 5 ] [ (x – 2) + √ 5 ] = (x – 3)[(x – 2) 2 – 5] Multiply. = (x – 3)[(x 2 – 4x + 4) – 5] Expand binomial. = (x – 3)(x 2 – 4x – 1) Simplify. = x 3 – 4x 2 – x – 3x 2 + 12x + 3 Multiply. = x 3 – 7x 2 + 11x + 3 Combine like terms.

3. EXAMPLE 3 Use zeros to write a polynomial function f(3) = 3 3 – 7(3) 2 + 11(3) + 3 = 27 – 63 + 33 + 3 = 0  f(2 + √ 5 ) = (2 + √ 5 ) 3 – 7(2 + √ 5 ) 2 + 11( 2 + √ 5 ) + 3 = 38 + 17 √ 5 – 63 – 28 √ 5 + 22 + 11 √ 5 + 3 = 0  Since f (2 + √ 5 ) = 0, by the irrational conjugates theorem f (2 – √ 5 ) = 0.  You can check this result by evaluating f (x) at each of its three zeros. CHECK

4. = (x + 1) (x 2 – 4x – 2x + 8) f (x) = (x + 1) (x – 2) ( x – 4) GUIDED PRACTICE Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros. 5. – 1, 2, 4 for Example 3 Write f (x) in factored form. = (x + 1) (x 2 – 6x + 8) Multiply. = x 3 – 6x 2 + 8x + x 2 – 6x + 8 = x3 – 5x 2 + 2x + 8 Multiply. Combine like terms. Use the three zeros and the factor theorem to write f(x) as a product of three factors. SOLUTION

5. GUIDED PRACTICE for Example 3 6. 4, 1 + √ 5 f (x) = (x – 4) [ x – (1 + √ 5 ) ] [ x – (1 – √ 5 ) ] Write f (x) in factored form. Regroup terms. = (x – 4) [ (x – 1) – √ 5 ] [ (x – 1) + √ 5 ] Multiply. = (x – 4)[(x 2 – 2x + 1) – 5] Expand binomial. = (x – 4)[(x – 1) 2 – ( 5) 2 ] Because the coefficients are rational and 1 + 5 is a zero, 1 – 5 must also be a zero by the irrational conjugates theorem. Use the three zeros and the factor theorem to write f (x) as a product of three factors SOLUTION

6. GUIDED PRACTICE for Example 3 = (x – 4)(x 2 – 2x – 4) Simplify. = x 3 – 2x 2 – 4x – 4x 2 + 8x + 16 Multiply. = x 3 – 6x 2 + 4x +16 Combine like terms.

7. GUIDED PRACTICE 7. 2, 2i, 4 – √ 6√ 6 for Example 3 Because the coefficients are rational and 2i is a zero, –2i must also be a zero by the complex conjugates theorem. 4 + 6 is also a zero by the irrational conjugate theorem. Use the five zeros and the factor theorem to write f(x) as a product of five factors. f (x) = (x–2) (x +2i)(x-2i)[(x –(4 – √ 6 )][x –(4+ √ 6) ] Write f (x) in factored form. Regroup terms. = (x – 2) [ (x 2 –(2i) 2 ][x 2 –4)+ √ 6][(x– 4) – √ 6 ] Multiply. = (x – 2)(x 2 + 4)(x 2 – 8x+16 – 6) Expand binomial. = (x – 2)[(x 2 + 4)[(x– 4) 2 – ( 6 ) 2 ] SOLUTION

8. GUIDED PRACTICE for Example 3 = (x – 2)(x 2 + 4)(x 2 – 8x + 10) Simplify. = (x–2) (x 4 – 8x 2 +10x 2 +4x 2 –3x +40) Multiply. = (x–2) (x 4 – 8x 3 +14x 2 –32x + 40) Combine like terms. = x 5 – 8x 4 +14x 3 –32x 2 +40x – 2x 4 +16x 3 –28x 2 + 64x – 80 = x 5 –10x 4 + 30x 3 – 60x 2 +10x – 80 Combine like terms. Multiply.

9. GUIDED PRACTICE 8. 3, 3 – i for Example 3 Because the coefficients are rational and 3 –i is a zero, 3 + i must also be a zero by the complex conjugates theorem. Use the three zeros and the factor theorem to write f(x) as a product of three factors = f(x) =(x – 3)[x – (3 – i)][x –(3 + i)] = (x–3)[(x– 3)+i ][(x 2 – 3) – i] Regroup terms. = (x–3)[(x – 3) 2 –i 2 )] = (x– 3)[(x – 3)+ i][(x –3) –i] Multiply. Write f (x) in factored form. SOLUTION

10. GUIDED PRACTICE for Example 3 = (x – 3)[(x – 3) 2 – i 2 ]=(x –3)(x 2 – 6x + 9) Simplify. = (x–3)(x 2 – 6x + 9) = x 3 –6x 2 + 9x – 3x 2 +18x – 27 Combine like terms. = x 3 – 9x 2 + 27x –27 Multiply.