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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.
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)[(x2 – 4x + 4) – 5]
Expand binomial.
= (x – 3)(x2 – 4x – 1)
Simplify.
= x3 – 4x2 – x – 3x2 + 12x + 3
Multiply.
= x3 – 7x2 + 11x + 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 + 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. 
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 the 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) (x2 – 4x – 2x + 8)
= (x + 1) (x2 – 6x + 8)
= x3 – 6x2 + 8x + x2 – 6x + 8
= x3 – 5x2 + 2x + 8
Write f (x) in factored form.
Multiply.
Combine like terms.
Multiply.
Combine like terms.
GUIDED PRACTICE
6.
for Example 3
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 the 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)[(x2 – 2x + 1) – 5]
Expand binomial.
GUIDED PRACTICE
for Example 3
= (x – 4)(x2 – 2x – 4)
Simplify.
= x3 – 2x2 – 4x – 4x2 + 8x + 16
Multiply.
= x3 – 6x2 + 4x +16
Combine like terms.
GUIDED PRACTICE
7.
for Example 3
2, 2i, 4 – √ 6
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.
SOLUTION
= (x – 2) [ (x2 –(2i)2][x2–4)+√6][(x– 4) – √6 ]
Write f (x) in
factored form.
Regroup terms.
= (x – 2)[(x2 + 4)[(x– 4)2 – ( 6 )2]
Multiply.
= (x – 2)(x2 + 4)(x2 – 8x+16 – 6)
Expand binomial.
f (x) = (x–2) (x +2i)(x-2i)[(x –(4 –√6 )][x –(4+√6) ]
GUIDED PRACTICE
for Example 3
= (x – 2)(x2 + 4)(x2 – 8x + 10)
Simplify.
= (x–2) (x4– 8x2 +10x2 +4x2 –3x +40) Multiply.
= (x–2) (x4 – 8x3 +14x2 –32x + 40)
Combine like terms.
= x5– 8x4 +14x3 –32x2 +40x – 2x4
+16x3 –28x2 + 64x – 80
Multiply.
= x5–10x4 + 30x3 – 60x2 +10x – 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 the 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 ][(x2 – 3) – i]
Write f (x) in factored
form.
Regroup terms.
= (x–3)[(x – 3)2 –i2)]
Multiply.
= (x– 3)[(x – 3)+ i][(x –3) –i]
GUIDED PRACTICE
for Example 3
= (x – 3)[(x – 3)2 – i2]=(x –3)(x2 – 6x + 9)
= (x–3)(x2 – 6x + 9)
Simplify.
= x3–6x2 + 9x – 3x2 +18x – 27
Multiply.
= x3 – 9x2 + 27x –27
Combine like terms.