# Order Polynomial and Rational Functions

Order Polynomial and Rational Functions
Order Polynomial and Rational Functions
(a) degree = 5; (b) domain = R, range = R; (c) zeros: x = –3 –2, 1, 2, 4; (d)
x-intercepts = –3, –2, 1, 2, 4; (e) y-intercept = p(0) = 3(0 – 1)(0 3)(0 – 4)(0 2)
(0 – 2) = –144
232. (a) degree = 6; (b) domain = R, range ⊂ R; (c) q(x) = (x
2 4)(x
2 – 5)(x
2 –
9) = (x 2i)(x – 2i)(x )(x – ) (x 3)(x – 3); zeros: x = –3, – , , 3,–2i,
2i; (d) x-intercepts = –3, – , , 3; (e) y-intercept = q(0) = (0
2 4)(0
2 – 5)(0
2 –
9) = 180
233. (a) degree = 4; (b) domain = R, range = [–81, ∞); (c) g(x) = x
4 – 81 =
(x
2 9)(x
2 – 9) = (x 3i)(x – 3i)(x 3)(x – 3); zeros: x = –3, 3,–3i, 3i; (d) xintercepts = –3, 3; (e) y-intercept = g(0) = 0
4 –81 = –81
234. (a) degree = 2; (b) domain = R, range = (–∞, 15.125] Note: The graph of g
is a parabola that turns downward with vertex (–2.25,15.125); (c) g(x) = –2x
2 –
9x 5 = –(2x
2 9x – 5) = –(2x – 1)(x 5); zeros: x = –5, ; (d) x-intercepts = –
5, ; (e) y-intercept = g(0) = –2 · 0
2 – 9 · 0 5 = 5
235. (a) degree = 1; (b) domain = R, range = R; (c) zero: x = – ; (d) x-intercept
= – ; (e) y-intercept = f (0) = 3 · 0 5 = 5
236. (a) turning points: (–7.22,40.31), (–2.42,–22.76), (1.97,3.41), (5.27,–
Order Polynomial and Rational Functions
17.68); (b) relative maxima: 40.31, 3.41; relative minima: –22.76, –17.68; no
absolute extrema
237. The remainder theorem states that if a polynomial p(x) is divided by x – a,
the remainder is p(a). Using synthetic division,
Thus, p(2) = –24.
238. Using synthetic division,
Thus, p(–2) = 0.
239. The factor theorem states that x – c is a factor of p(x) if and only if p(c) =
0. Therefore, because p(–2) = 0, (x 2) is a factor of p(x). From the synthetic
division in question 238 p(x) = (x 2)(2x
2 – 9x 4). Factoring the quadratic
yields p(x) = (x 2)(2x
2 – 9x 4) = (x 2)(2x – 1)(x – 4).
240. p(x) = 5(x – 3)(x 2)(x )(x – )
241. g(x) = 2(x 1)(x – 1)(x – 3)
242. (A) one
(B) k
(C) roots
(D) n
(E) x – yi
243. (a) p(x) = (x 3)(x
2 – 5)(x
2 – 49)(x
2 49)
= (x 3)(x )(x – )(x 7)(x – 7)(x 7i)(x – 7i)
(b) zeros: 7i, –7i, –7, –3, – , , 7
244. (a) p(x) = (3x
2 – x – 10)(x
6 – 64) = (3x 5)(x – 2)(x
3 8)(x
3 – 8)
= (3x 5)(x – 2)(x 2)(x
2 – 2x 4)(x – 2)(x
2 2x 4)
= (3x 5)(x – 2)(x 2)(x 1 i )(x 1 – i )(x – 2)(x –1 i )(x – 1 – i
)
= (3x 5)(x – 2)
2 (x 2)(x 1 i )(x 1 – i )(x – 1 i )(x – 1 – i )
(b) zeros: – , 2 (multiplicity 2), –2, 1 i , 1 – i , –1 i , –1 – i )
245. The polynomial has real coeffcients; so another root is 2 i, the complex
conjugate of 2 – i. Each root corresponds to a factor of the polynomial yielding
p(x) = (x – 3)[x – (2 – i)][x – (2 i)]
p(x) = (x – 3)[(x – 2) i][(x – 2) – i] Regroup.
p(x) = (x – 3)[(x – 2)
2 – i
2] Multiply last two factors.
p(x) = (x – 3)[(x
2 – 4x 4) – (–1)] Simplify.
p(x) = x
3 – 7x
2 17x – 15
Order Polynomial and Rational Functions
This polynomial equation has the required roots and the least degree.
246. The number of sign variations in p(x) = 6x
4 7x
3 – 9x
2 – 7x 3 is 2, so
p(x) = 0 has 2 or 0 positive real roots. The number of sign variations in p(–x) =
6x
4 – 7x
3 – 9x
2 7x 3 is 2, so p(x) = 0 has 2 or 0 negative real roots.
247. The number of sign variations in p(x) = x
3 – 1 is 1, so p(x) = 0 has at most
one positive real root. The number of sign variations in p(–x) = –x
3 –1 is 0, so
p(x) = 0 has no negative real roots.
248. The number of sign variations in p(x) = x
5 4x
4 – 4x
3 – 16x
2 3x 12 is
2, so p(x) = 0 has 2 or 0 positive real roots. The number of sign variations in
p(–x) = –x
5 4x
4 4x
3 –16x
2 – 3x 12 is 3, so p(x) = 0 has 3 or 1 negative real
roots.
249. The rational root theorem states that if anx
n an–1x
n–1 an–2x
n–2 …
a2x
2 a1x a0 = 0 is a polynomial equation with integral coeffcients and it has a
rational root (in lowest terms), then p is a factor of a0 and q is a factor of an
.
Possible numerators of a rational zero are factors of 6: ±1, ±2, ±3, and ±6.
Possible denominators are factors of 2: ±1 and ±2. Thus, possible rational roots
of 2x
3 x
2 – 13x 6 = 0 are ±1, ±2, ±3, ±6, ± , and ± .
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