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Section%203.4

# Section%203.4 - 6 “may 0 MA 180 — Precalculus Professor...

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Unformatted text preview: /' ' 6 “may 0%- MA 180 — Precalculus Professor Terry Section 3.4: Complex and Rational Zeros of Polynomials E Complex Zeros Theorem on Conjugate Pair Zeros of a Polynomial: If a polynomial f (x) of degree n >1 has real Coefﬁcients and if z : a +bi with b at 0 is a complex zero of f (x) , then the conjugate Eza—biis also azero of f(x). 4 ﬁndg polynomial f (x) with real coefﬁcients and leading coefﬁcient 1 that has zero<—§ and f _ '\ ‘—"—_-_‘—'_'_‘"— , V‘ LYL (1—71 nddegreeB. “:2; dpﬁW33 {mﬂ63_3‘ if“ I (+7L - 59a} = (1(x4cXX—c—JJCM—Ce) . _ §§§_3__=ctC__><-f (§3?]__,Cf7(_ltt_1_i)] gm % l (mzﬂUﬂUCx-l-mg T polynomial with real coefﬁcients and positive degree n can be expressed as a product of linear and quadratic polynomials with real coefﬁcients such that the quadratic factors are irreducible over SR. Proof: Since f(x) have precisely n complex zeros c1, 02, C", we can write f(x) = 8(x — 01)(X - C2) . . - (x — on). where a is the leading coefﬁcient of f(x). Of course, some of the zeros may be real. In such cases, we obtain the linear factors referred to in the statement. If a zero ck is not real, then by the theorem of conjugate pair zeros of a polynomial, the conjugate Ek is also a zero of f(x) and hence most be one of the numbers 01, c2, cn_This implies that both x — ck and x - Ek appear in the factorization of f(x) if those factors are multiplied, we obtain (x_ck)(x—Ek) = ’52 ‘(ck +Ek)x+ckak) which has real coefﬁcients since ck +Ek and CE,( are real numbers. Thus, if ck is a complex zero, then the product (x—ck)(x—Ek) is a quadratic polynomial that is irreducible over SR. This completes the proof. MA180 Sec3.4 SpringO3 ewt Express the polynomial as a product of linear and quadratic factors: x5 —4x3 —x1 +4 :‘Sewixzqu-l 2 XBKQQH (35:43 , m m_jlégtq>(£éfi>_t_ ,,,,, _W * 4*“???le E'Céfié?([email protected]"m‘2“@ Rational Zeros Theorem on Rational Zeros of a Pol nomial: if the polynomial has integer coefﬁcients and if did is a rational zero of f(x) such that c and d have no common prime factor, then 1) the numerator c of the zero is a factor of the common term a; 2) the denominator d of the zero is a factor of the leading coefﬁcient an Show that the equatioréla —4x2 + 7x @ O has no rational root. . s3 quOLQLL «14/ hwmwcajgvA/I C 2 :t:[ ig ) . W‘QAL CitﬂmeLMEES-K/ é 212 I '€§ﬁ%%::£%&gﬂ i4“ .~ A /“ ,il [email protected] f 48S \$0) t O - 953* 0 it L/ a? 9 MA180 Sec3.4 Spring03 ewt r x . . 2. r - ¥on2 Bygﬁqx +7y+gf Finding the Rational Solutions of an Eguation Find all solutions of the equation 3 +83:2 -—[email protected] 0. J+o<2 (355523 —-i (and?) :4 O Use a calculator to ﬁnd the rational solutions for 12::3 + 8x2 7- 3s— 2 = 0. . {1 _ AL} M 6 Constructing a crate: The frame for a shipping crate is to be constructed from 24 feet of 2 X 2 lumber. Assuming that crate is to have square ends of length x feet, determine the value(s) of x that result(s) in a volume of .4 ft3. V W a V .3 H gﬁ 3 F’J am 9% 09 93a? M ’é . W MA180 Sec3.4 SpringD3 ewt X 3 - Co K 2+ L} 5: O ...
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Section%203.4 - 6 “may 0 MA 180 — Precalculus Professor...

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