Given that #f(x)=x^3+4x^2+bx+c#. When f is divided by #(x-3)#, the remainder is 110. In addition, when f is divided by #(x+2)#, the remainder is 150. The sum of #b+c=?#

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Answer 1 · 2017-10-01T04:53:31

Given that #f(x)=x^3+4x^2+bx+c#. When f is divided by #(x-3)#, the remainder is 110.
So we have #f(3)=110#

#=>3^3+4*3^2+3b+c=110#

#=>3b+c=47......[1]#

In addition, when f is divided by #(x+2)#, the remainder is 150.

This means

#f(-2)=150#

#=>(-2)^3+4*(-2)^2-2b+c=150#

#=>-2b+c=142.......[2]#

Subtracting [2] from [1] we get

#5b=-95#

#=>b=-19#

Inserting #b=-19# in [1] we get

#3*(-19)+c=47#

#=>c=104#

So the sum of #b+c=104-19=85#

Given that #f(x)=x^3+4x^2+bx+c#. When *f* is divided by #(x-3)#, the remainder is 110. In addition, when *f* is divided by #(x+2)#, the remainder is 150. The sum of #b+c=?#