2.5 Practice
Find an nth degree polynomial function with real coefficients satisfying the given conditions. If
you are using a graphing utility, use it to graph the function and verify the real zeros and the
given function value.
1) n = 3; 2 and 4i are zeros; f(-1) = -102
f(x) = a(x – 2)(x – 4i)(x + 4i)
f(x) = a(x – 2)(x2 + 4ix – 4ix – 16i2)
f(x) = a(x – 2)(x2 + 16)
f(-1) = a(-1 – 2)((-1)2 + 16) = a(-3)(17) = -102
a(-51) = -102,
a=2
f(x) = 2(x – 2)(x2 + 16)
If we foil things out and then distribute
f(x) = 2(x3 + 16x – 2x2 – 32)
f(x) = 2x3 – 4x2 + 32x – 64
2) n = 4; 2, -2, and 2i are zeros; f(0) = 64
f(x) = a(x – 2)(x + 2)(x – 2i)(x + 2i)
f(x) = a(x2 – 4)(x2 + 2ix – 2ix – 4i2) = a(x2 – 4)(x2 – 4i2)
f(x) = a(x2 – 4)(x2 + 4)
f(0) = a(02 – 4)(02 + 4) = a(-4)(4) = 64
-16a = 64
a = -4
f(x) = -4(x2 – 4)(x2 + 4)
If we foil things out and then distribute
f(x) = -4(x4 + 4x2 – 4x2 – 16) = -4(x4 – 16)
f(x) = -4x4 + 64
For each problem complete each of the following steps
a) List all of the possible rational roots
b) Use your calculator to find one real zero
c) Use synthetic division to verify it is zero of the polynomial (MANDATORY)
d) Factor the rest of the polynomial/use quadratic formula
e) Give all the zeros
1) f(x) = 2x3 + 6x2 + 5x + 2
2) f(x) = x3 + x2 – 4x – 4
a) ± 1, 1/2, 2
a) ± 1, 2, 4
b) x = -2 from the calculator
b) x = 2 from the calculator
c)
c)
(x – 2)(2x2 + 2x + 1)
−2 ± 4 − 4 ⋅ 2 ⋅1 −2 ± −4
x=
=
2
⋅
2
4
d)
−2 ± 2i #
1 1 &
x=
% or − ± i (
$
4
2 2 '
e) x = -2,
−2 + 2i −2 − 2i
,
4
4
(x – 2)(x2 + 3x + 2)
d) (x – 2)(x + 2)(x + 1)
e) x = 2, x = -2, x = -1