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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=?#

Answer 1

Christian Antunez

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#

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Answer 2

Xavier Alonso

The sum of b + c is 464.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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=?#

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=?#