Question #262421

(a) Find the polynomial degree 3 in term of three linear factors if P(1) = P(-2) = 0,



P(3) = 200 and P(-1) = 8.



(b) Find the partial fraction for



12x³ + 1/P(x)

1
Expert's answer
2021-11-09T16:29:06-0500

a)

Let P3(x)=a(xx1)(xx2)(xx3).P_3(x)=a(x-x_1)(x-x_2)(x-x_3).

Given x1=1,x2=2.x_1=1, x_2=-2. Then


P3(x)=a(x1)(x+2)(xx3)P_3(x)=a(x-1)(x+2)(x-x_3)

P(3)=200P(3)=200


P3(3)=a(31)(3+2)(3x3)=200P_3(3)=a(3-1)(3+2)(3-x_3)=200

a(3x3)=20a(3-x_3)=20

P(1)=8P(-1)=8


P3(1)=a(11)(1+2)(1x3)=8P_3(-1)=a(-1-1)(-1+2)(-1-x_3)=8

a(1x3)=4a(-1-x_3)=-4

3x31x3=204\dfrac{3-x_3}{-1-x_3}=\dfrac{20}{-4}

3x3=5+5x33-x_3=5+5x_3

x3=13x_3=-\dfrac{1}{3}

a(1(13))=4a(-1-(-\dfrac{1}{3}))=-4

a=6a=6

P3(x)=6(x1)(x+2)(x+13)P_3(x)=6(x-1)(x+2)(x+\dfrac{1}{3})

=(6x+2)(x2+x2)=(6x+2)(x^2+x-2)

=6x3+6x212x+2x2+2x4=6x^3+6x^2-12x+2x^2+2x-4

Answer:

P(x)=6x3+8x210x4P(x)=6x^3+8x^2-10x-4



b)


12x3+1P(x)=12x3+16x3+8x210x4\dfrac{12x^3+1}{P(x)}=\dfrac{12x^3+1}{6x^3+8x^2-10x-4}

=A6+Bx1+Cx+2+Dx+13=\dfrac{A}{6}+\dfrac{B}{x-1}+\dfrac{C}{x+2}+\dfrac{D}{x+\dfrac{1}{3}}

A(x1)(x+2)(x+13)+6B(x+2)(x+13)A(x-1)(x+2)(x+\dfrac{1}{3})+6B(x+2)(x+\dfrac{1}{3})

+6C(x1)(x+13)+6D(x1)(x+2)+6C(x-1)(x+\dfrac{1}{3})+6D(x-1)(x+2)

=12x3+1=12x^3+1

x=1:6B(3)(43)=12+1=>B=1324x=1:6B(3)(\dfrac{4}{3})=12+1=>B=\dfrac{13}{24}

x=2:6C(3)(53)=96+1=>C=196x=-2: 6C(-3)(-\dfrac{5}{3})=-96+1=>C=-\dfrac{19}{6}

x=13:6D(43)(53)=49+1=>D=124x=-\dfrac{1}{3}: 6D(-\dfrac{4}{3})(\dfrac{5}{3})=-\dfrac{4}{9}+1=>D=-\dfrac{1}{24}

x=0:23A+23(134)13(19)2(14)=1x=0:-\dfrac{2}{3}A+\dfrac{2}{3}(\dfrac{13}{4})-\dfrac{1}{3}(-19)-2(-\dfrac{1}{4})=1

A=12A=12

Answer:

12x3+16x3+8x210x4=2+1324(x1)\dfrac{12x^3+1}{6x^3+8x^2-10x-4}=2+\dfrac{13}{24(x-1)}

196(x+2)124(x+13)-\dfrac{19}{6(x+2)}-\dfrac{1}{24(x+\dfrac{1}{3})}


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