Maths....Polynomial with ans－dvhtlbz的部落格

標題:

Maths....Polynomial with ans

此文章來自奇摩知識+如有不便請留言告知

發問:

http://x7c.xanga.com/d24c924026c31221313566/b173611045.jpg can u plz show me how to do Q4, Q5(b) ,and Q7-10 thanks so much~~x] 更新: ANSWERS::: Q4 a = 1, b = -3, c = -6 Q5(b) P(x) = (x+1)(x-4)^2 更新 2: ************** no need to do Q7-9.............only Q4, Q5(b), Q10(b) thz~ 更新 3: actually the answer for 10 b) is u=25; -v=54 although no one got that right,, but thankyou so much for ur time and effort

最佳解答:

Q4) Given a polynomail: P(x)=ax^3-bx^2+cx-8 and 3 conditions: 2 is a zero , that means P(2) =0 ...(1) -1 is a zero , that means P(-1)=0 ...(2) P(3)=28 ...(3) from (1) we have a(2)^3-b(2)^2+c(2)-8=0 8a-4b+2c-8=0 4a-2b+c=4 ... (4) from (2) we have a(-1)^3-b(-1)^2+c(-1)-8=0 -a-b-c-8=0 a+b+c=-8 ... (5) from (3) we have a(3)^3-b(3)^2+c(3)-8=28 27a-9b+3c-8=28 27a-9b+3c=36 9a-3b+c=12...(6) from (5) we know c=-a-b-8...(7) sub (7) to (4) and (6) respectively (4) becomes: 4a-2b+(-a-b-8)=4 4a-2b-a-b-8=4 3a-3b=12 a-b=4 ...(8) (6) becomes: 9a-3b+(-a-b-8)=12 9a-3b-a-b-8=12 8a-4b=20 2a-b=5... (9) solving (8) and (9) simultaneously (9)-(8): a=1 ...(10) sub (10) into (8) 1-b=4, so b=-3 ...(11) sub (10) and (11) into (7), c=-(1)-(-3)-8 =-6 Q5) (b) because we know 4 is a root and actually, 4 is a repeated root hence (x-4) and (x-4) are two factors of the polynomial by looking at the constant term of the polynomial: 16 we know the remaining factor is something like (x+1) then we calculate P(-1), it happens to be 0 hence, by factor theorm, (x+1) is another factor so P(x) = (x+1)(X-4)^2 10) (b) let P(x) = x^5-9x^4+ux^3-18x^2+v+27 differentiate P(x) w.r.t. x P'(x)= 5x^4 -36x^3+3ux^2-36x we have two given condtion: P(3)=0 ...(1) P'(3)=0 ...(2) consider (2) first 5(3)^4 -36(3)^3 +3u(3)^2 -36(3)=0 405 - 324 +27u -108=0 u=1 ...(3) consider (1) and sub. (3) into it (3)^5-9(3)^4+(1)(3)^3-18(3)^2+v+27=0 324 -729 +27 -162 +v +27=0 v=513

其他解答:

4) zeros mean roots P(2) = 8a - 4b ＋ 2c - 8 = 0 P(-1) = -a - b - c - 8 = 0 P(3) = 27a - 9b ＋ 3c = 28 then sovle the 3 equ 3 unknown by any method you like such as elimination 5(b) no need step, just write down the answer you may try to do division to get the other factor i.e. x ＋ 1 --------------------------------- (x^2 - 8x＋16) / x^3 - 7x^2 ＋ 8x ＋ 16 -) x^3 - 8x^2 ＋ 16x ________________________ x^2-8x ＋ 16 x^2-8x ＋ 16 ________________________ 10b) let P(x) = x^5 - 9x^4 ＋ux^3 - 18x^2 ＋ v＋27 given P(x) = 0 has a triple root at x =3, => P(3) = 0 and P'(3) = 0 P'(3)=0 => 405 ＋ 972＋ 27u -108 =0 => u=-47 P(3) = 0 =>243 - 729 ＋ 27u - 162 ＋ v 27 =0 => v = 621-27u = 18907638E748CCC65837