if x cube + ax square- bx + 10 is divided by x square - 3 x + 2 find the value of a and b
Answers
Answer:
it can be solved by two method
Step-by-step explanation:
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 . Answer
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