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