Answer to Question #262421 in Math for Shah

Question

(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)

Answer 1

a)

Let $P_3(x)=a(x-x_1)(x-x_2)(x-x_3).$

Given $x_1=1, x_2=-2.$ Then

$$P_3(x)=a(x-1)(x+2)(x-x_3)$$

$P(3)=200$

$$P_3(3)=a(3-1)(3+2)(3-x_3)=200$$

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

$P(-1)=8$

$$P_3(-1)=a(-1-1)(-1+2)(-1-x_3)=8$$

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

$$\dfrac{3-x_3}{-1-x_3}=\dfrac{20}{-4}$$

$$3-x_3=5+5x_3$$

$$x_3=-\dfrac{1}{3}$$

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

$$a=6$$

$$P_3(x)=6(x-1)(x+2)(x+\dfrac{1}{3})$$

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

$$=6x^3+6x^2-12x+2x^2+2x-4$$

Answer:

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

b)

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

$$=\dfrac{A}{6}+\dfrac{B}{x-1}+\dfrac{C}{x+2}+\dfrac{D}{x+\dfrac{1}{3}}$$

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

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

$$=12x^3+1$$

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

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

$$x=-\dfrac{1}{3}: 6D(-\dfrac{4}{3})(\dfrac{5}{3})=-\dfrac{4}{9}+1=>D=-\dfrac{1}{24}$$

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

$$A=12$$

Answer:

$$\dfrac{12x^3+1}{6x^3+8x^2-10x-4}=2+\dfrac{13}{24(x-1)}$$

$$-\dfrac{19}{6(x+2)}-\dfrac{1}{24(x+\dfrac{1}{3})}$$