Question

If $x^3+a{x}^2+bx+6$ has $x-2$ as a factor and leaves a remainder 3 when divided by $x-3$. Find the values of a and b.

• A. a = 3 and b = 1
• B. a = -3 and b = 1
• C. a = -3 and b = -1
• D. a = 3 and b = -1

Answer 1

The correct option is C

a = -3 and b = -1

$f\left(x\right)=x^3+a{x}^2+bx+6$

Since $x-2$ is a factor, $f\left(2\right)=0$

$(2{)}^3+a(2{)}^2+b\left(2\right)+6=0$

Simplyfying we get $2a+b=-7$ ........(i)

Since $x-3$ leaves a remainder 3, $f\left(3\right)=3$

$(3{)}^3+a(3{)}^2+b\left(3\right)+6=3$

Simplyfying we get $3a+b=-10$ ........(ii)

Solving (i) and (ii) we get

a = -3 and b = -1