Question

The polynomial a$x^3$ + b$x^2$ + x – 6 has (x + 2) as a factor and leaves a remainder 4 when divided by (x – 2). Find a and b.

Answer 1

Let p(x)= a$x^3$ + b$x^2$ + x -6

By using factor theorem, (x+2)is a factor of p(x), only when p(-2) = 0
p(-2) = a${(-2)}^3$ + b${(-2)}^2$ + (-2)-6=0

$\Rightarrow$ -8a+4b-8=0

$\Rightarrow$ -2a+b=2 ...(i)

Also when p(x) is divided by (x- 2) the reminder is 4.

∴ p(2)=4

$\Rightarrow$ a${\left(2\right)}^3$+b${\left(2\right)}^2$+2-6=4

$\Rightarrow$ 8a+4b+2-6=4

$\Rightarrow$ 8a+4b=8

$\Rightarrow$2a+b=2 ...(ii)

Adding equation(i) and (ii) weget(-2a+b)+(2a+b)=2+2

$\Rightarrow$​ 2b=4

$\Rightarrow$ b=2

Putting b=2 in (i) we get

$\Rightarrow$-2a+2=2

$\Rightarrow$-2a=0

$\Rightarrow$a=0
Hence,a=0 and b=2