SOLUTION: Consider the polynomial F(x)=4x^3+(a+1)x^2+x-5b and G(x)=4x^3-ax^2+bx-2, where a and b are constants. When both F(x) and G(x) are divided by x-1, the remainders are 3 and -3 respec

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Question 1200512: Consider the polynomial F(x)=4x^3+(a+1)x^2+x-5b and G(x)=4x^3-ax^2+bx-2, where a and b are constants. When both F(x) and G(x) are divided by x-1, the remainders are 3 and -3 respectively.
a)find the values of a and b
b)Solve the equation F(x)-G(x)=-2

Answer by ikleyn(52792)   (Show Source): You can put this solution on YOUR website!
.
Consider polynomials F(x) = 4x^3+(a+1)x^2+x-5b and G(x) = 4x^3-ax^2+bx-2,
where a and b are constants.
When both F(x) and G(x) are divided by x-1, the remainders are 3 and -3, respectively.
a) Find the values of a and b.
b) Solve the equation F(x)-G(x) = -2.
~~~~~~~~~~~~~~~~~~~~~


                Step by step solution


(a)  We are given that F(x) = 4x^3+(a+1)x^2+x-5b gives the remainder 3 when divided by (x-1).

     According to the Remainder theorem, it means that  F(1) = 3.

     Calculate F(1) by substituting x= 1 into the formula for F(x)

        F(1) = 4*1 + (a+1)*1 + 1 - 5b = 4 + a+1 + 1 - 5b = a - 5b + 6.


     Hence, we have THIS equation for "a" and "b"

        a - 5b + 6 = 3,   or  a - 5b = -3.    (1)



(b)  Next, we are given that G(x) = 4x^3-ax^2+bx-2 gives the remainder -3 when divided by (x-1).

     According to the Remainder theorem, it means that  G(1) = -3.

     Calculate G(1) by substituting x= 1 into the formula for G(x)

        G(1) = 4*1 - a*1 + b*1 - 2 = 4 - a + b - 2 = -a + b + 2.


     Hence, we have THIS equation for "a" and "b"

        -a + b + 2 = -3,   or  -a + b = -5.    (2)



(c)  Thus we have two equations to find "a" and "b"

        a - 5b = -3    (1)
       -a +  b = -5    (2)


     To solve, add the equations

           -4b = -8  ===>  b = (-8)/(-4) = 2.


     Then from equation (1),

        a = -3 + 5b = -3 + 5*2 = -3 + 10 = 7.


     +--------- ANSWER------------+
     |     Thus  a = 7;  b = 2.   |
     +----------------------------+


First part is complete.

----------------

The second part, after subtracting polynomials, gives a quadratic polynomial .
Working with it is simple arithmetic, so I leave this part on you.



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