EXAMPLE 3 Use zeros to write a polynomial

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 three zeros and the factor theorem to write f (x) as a product of three factors.

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. = (x – 3) [ (x – 2) – √ 5 ] [ (x – 2) +√ 5 ] Regroup terms. = (x – 3)[(x – 2)2 – 5] Multiply. = (x – 3)[(x 2 – 4 x + 4) – 5] Expand binomial. = (x – 3)(x 2 – 4 x – 1) Simplify. = x 3 – 4 x 2 – x – 3 x 2 + 12 x + 3 Multiply. = x 3 – 7 x 2 + 11 x + 3 Combine like terms.

EXAMPLE 3 Use zeros to write a polynomial function CHECK You can check this result by evaluating f (x) at each of its three zeros. f(3) = 33 – 7(3)2 + 11(3) + 3 = 27 – 63 + 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.

GUIDED PRACTICE for Example 3 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 Use three zeros and the factor theorem to write f(x) as a product of three factors. SOLUTION f (x) = (x + 1) (x – 2) ( x – 4) = (x + 1) (x 2 – 4 x – 2 x + 8) = (x + 1) (x 2 – 6 x + 8) = x 3 – 6 x 2 + 8 x + x 2 – 6 x + 8 = x 3 – 5 x 2 + 2 x + 8 Write f (x) in factored form. Multiply. Combine like terms.

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

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

GUIDED PRACTICE 7. for Example 3 2, 2 i, 4 – √ 6 Because the coefficients are rational and 2 i is a zero, – 2 i 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. SOLUTION = (x – 2) [ (x 2 –(2 i)2][x 2– 4)+√ 6][(x– 4) – √ 6 ] Write f (x) in factored form. Regroup terms. = (x – 2)[(x 2 + 4)[(x– 4)2 – ( 6 )2] Multiply. = (x – 2)(x 2 + 4)(x 2 – 8 x+16 – 6) Expand binomial. f (x) = (x– 2) (x +2 i)(x-2 i)[(x –(4 –√ 6 )][x –(4+√ 6) ]

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

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

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