Question

when f(x)= x^4+3x^3+px^2-2x+q is divided by (x-2) the remainder is 15, and (x+3) is a factor of f(x). find values of p and q solutions

Answer 1

When f(x)= x^4+3x^3+px^2-2x+q is divided by (x-2) the remainder is 15, and (x+3) is a factor of f(x). find values of p and q

Division by (x-2). Remainder = 15.

        

(x-2) ) x^4+3x^3+px^2 - 2x + q( x^3 + 5x^2 + (10+p)x + (18+2p)

           x^4-2x^3

                  5x^3+px^2

                  5x^3-10x^2

                           (10+p)x^2 – 2x

                           (10+p)x^2 – (20+2p)x

                                               (18+2p)x + q

                                               (18+2p)x – (36+4p)

                                                                 (36+4p+q)

Remainder is R1 = 36 + 4p + q = 15

 

Division by (x+3). Remainder = 0, since (x+3) is a factor

        

(x+3) ) x^4+3x^3+px^2 - 2x + q( x^3 + px – (3p+2)

           x^4+3x^3

                    0  + px^2 – 2x

                           px^2 + 3px

-(3p+2)x + q

                                      -(3p+2)x – (9p+6)

                                                        9p+q+6

Remainder is R2 = 9p + q + 6 = 0

Using R1 and R2,

4p + q = -21

9p + q = -6

Subtracting,

5p = 15

p = 3

q = -33

Answer: the values of p and q are: p = 3, q = -33

Answer 2

If we subtract the remainder we get x-2 is a factor of x⁴+3x³+px²-2x+q-15.

Using synthetic division we see that q+4p+21=0. Therefore, q=-4p-21.

We also know that x+3 is a factor of the original equation and using synthetic division we see that q+9p+6=0. Therefore, q=-9p-6. Equating the two expressions for a we get -4p-21=-9p-6 or we can write this as 4p+21=9p+6 and solve for p: 15=5p, p=3, making q=-33.

The original equation becomes:

x⁴+3x³+3x²-2x-33.