Question

X3+ax2-bx+10 is divisible by x2-3x+2,fund the values of a and b

Answer 1

. First method:-

x^3+ax^2-bx+10

=x(x^2–3x+2)+3x^2–2x+ax^2-bx+10

=x(x^2–3x+2)+(3+a)x^2-(2+b)x+10

=x(x^2–3x+2)+5[(3+a)/5.x^2-(2+b)/5.x+2]

If x^3+ax^2-bx+10 is divisible by x^2–3x+2 then

(3+a)/5.x^2-(2+b)/5.x+2=x^2–3x+2

by equating the coeff.of x^2 and x .

(3+a)/5=1 or 3+ a= 5 => a= 2 , Answer

(2+b)/5=3 or 2+b=15 => b=13 , Answer

Second method:-

Divisor= x^2–3x+2 =0

( x- 2) (x - 1) =0

x= 2 , 1 .

put x=2 , remainder=(2)^3+a(2)^2 -b(2) + 10=0

8+4a-2b+10=0 or 4a-2b= -18

2a-b = -9………………..(1)

put x = 1 , remainder = 1^3+a.1^2-b.1+10=0

1+a-b+10=0 , or a - b = -11……………(2)

subtract eq.(2) from (1)

a = 2 , put a =2 in eq. (2)

2-b = -11

2+11 = b

b = 13

a = 2 , b = 13 .