Question

find the sum of the remainder when the polynomial p(x)=x³-3x²+4x-9 is divided by(x-1)(x+2)​

Answer 1

Answer:

p(x)=x³-3x²+4x-9

by remainder theorem,

put x-1=0

x=1

hence, p(1) is remainder

remainder(r)=p(1)=(1)³-3(1)²+4(1)-9

p(1)=1-3+4-9

=-7

now,

put,

x+2=0

x=-2

p(-2) is remainder when p(x) divided by x+2

remainder(R)=p(-2)= (-2)³-3(-2)²+4(-2)-9

=-8-12-8-9

=-37

sum= -37+(-7)

=-44

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Step-by-step explanation:

Answer 2

Answer:

p(1)=-3

p(-2)=-46

Step-by-step explanation:

p(x)=x³-3x²+4x-9

g(x)=(x-1)

Now,find the zero of g(x)

x-1=0

x=1

Put x=1 in p(x)

p(1)=1³*-3*1²+4*1-9

p(1)=(1-3)+(4-9)

p(1)=-2+(-5)

p(1)=-2-5

p(1)=-7. [equation 1]

Now,find the zero of g(x)

x+2=0

x=-2

Put x=-2 in p(x)

p(-2)=-2³*-3*-2²+4*-2-9

p(-2)=-8-3*4+4*-2-9

p(-2)=-8-12+(-8)-9

p(-2)=-20+(-17)-9

p(-2)=(-20-17)-9

p(-2)=-37-9

p(-2)=-46. [equation 2

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