Question

On dividing 2x³-5x²+1 by g(x), if remainder is 17x-6 and quotient is 2x-7 , find g(x)​

Answer 1

Answer:

as the quotient has degree 1 then the degree of gx must be 2

f(x)=g(x)×q(x)+r(x)

$2 {x}^{3} - 5 {x}^{2} + 1 = a {x}^{2} + bx + c \times (2x - 7) + 17x - 6 \\ 2{x}^{3} - 5 {x}^{2} + 1 = 2a {x}^{3} - 7a {x}^{2} + 2b {x}^{2} - 7bx + 2cx - 7c + 17x - 6 \\ 2 {x}^{3} - 5 {x}^{2} + 1 = 2a {x}^{3} + {x}^{2} ( - 7a + 2b) + x( - 7b + 17) - 7c - 6$

comparing the terms with the terms of same degree

2=2a. ,. -5=-7a+2b. 1=-7c+1

a=1. -5=-7+2b. c=0

-5+7=2b

b=1

gx is

$1 {x}^{2} + 1x + 0$