Question

if if p(x)=x4+7x3+7x2++px+q is exactly divisible by x2+4x+3 find p and q​

Answer 1

Answer:

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Answer 2

x2 +4x+3

x2+3x+x+3

x(x+3)+1(x+3)

(x+3)(x+1).

g(x)=x+3

g(0)=x+3=0

x=_3.

p(x)=x4+7x3+7x2+px+q =0

p(_3)=(_3)4+7×(_3)2+7×(_3)3+p×_3+q=0

=81+63_189_3p+q=0

= 81+63_3p+q=0

= 81_3p+q=0

= _3p+q=_81.in equation first.

g(x)=x+1

g(0)=x+1=0

=x=_1.

p(_1)=(_1)4+7×(_1)3+7×(_1)2+p×_1+q=0

=1_7+7_p+q=0

=1_p+q=0

=_p+q=_1.in equation second.

by subtracting equation 1 from 2

_2p=_80

_p=_80÷2

_p=_40

p=40

by putting equation 1.

_3p+q =_81

_3×40+q=_81

_120+q=_81

q=_81+120

q=39.

p=40,q=39.