Question

If $a{x}^3+b{x}^2+x-6$ leves a remainder 0 and 4 when divided by x + 2 and (x - 2) respectively, find the values of a and b.

Answer 1

Let $p\left(x\right)=a{x}^3+b{x}^2+x-6$ be the given polynomial.
It is given that p(x) leaves the remainder 0 when it is divided by (x + 2). Therefore,
p(-2) = 0 $[x+2=0\Rightarrow x=-2]$
$\Rightarrow a(-2{)}^3+b(-2{)}^2+(-2)-6=0$
$\Rightarrow -8a+4b-2-6=0\Rightarrow -8a+4b=8$
$\Rightarrow$ - 2a + b = 2 ..... (i)
It is given that p(x) leaves the remainder 4 when it is divided by (x - 2). Therefore, p(2) = 4 $[x-2=0\Rightarrow x=2]$
$\Rightarrow$ $a(2{)}^3+b(2{)}^2+2-6=4$
$\Rightarrow$ 8a + 4b - 4 = 4 $\Rightarrow$ 8a + 4b = 8
$\Rightarrow$ 2a + b = 2 .... (ii)
Adding (i) and (ii), we get
2b = 4 $\Rightarrow$ b = 2
Putting b = 2 in (i), we get
-2a + 2 = 2 $\Rightarrow$ -2a = 0 $\Rightarrow$ a = 0
Hence, a = 0 and b = 2.