Question

if 2x cube+ax square-11x+b leaves a remainder of 0 and 42 when divided by (x-2) and (x-3) respectively, find a and b​

Answer 1

Let:

f(x)=x3+ax2+bx+6fx=x3+ax2+bx+6

(x−2) is a factor of f(x)=x3+ax2+bx+6.⇒f(2)=0⇒23+a×22+b×2+6=0⇒14+4a+2b=0⇒4a+2b=−14⇒2a+b=−7         ...(1)x-2 is a factor of fx=x3+ax2+bx+6.⇒f2=0⇒23+a×22+b×2+6=0⇒14+4a+2b=0⇒4a+2b=-14⇒2a+b=-7         ...1

Now, 

x−3=0⇒x=3x-3=0⇒x=3

By the factor theorem, we can say:

When f(x) will be divided by (x−3), 3 will be its remainder.⇒f(3)=3When fx will be divided by x-3, 3 will be its remainder.⇒f3=3

Now,

f(3)=33+a×32+b×3+6     =(27+9a+3b+6)     =33+9a+3bf3=33+a×32+b×3+6     =27+9a+3b+6     =33+9a+3b

Thus, we have:

    f(3)=3⇒33+9a+3b=3⇒9a+3b=−30⇒3a+b=−10      ...(2)    f3=3⇒33+9a+3b=3⇒9a+3b=-30⇒3a+b=-10      ...2

Subtracting 11 from 22, we get:

a = −-3

By putting the value of a in 11, we get the value of b, i.e., −-1.

∴ a = −-3 and b = −-1