Using the remainder theorem, how do you find the remainder of 3x^5-5x^2+4x+1 when it is divided by (x-1)(x+2)?

Nov 8, 2017

$42 x - 39 = 3 \left(14 x - 13\right) .$

Explanation:

Let us denote, by $p \left(x\right) = 3 {x}^{5} - 5 {x}^{2} + 4 x + 1 ,$ the given

polynomial (poly.).

Noting that the divisor poly., i.e., $\left(x - 1\right) \left(x + 2\right) ,$ is of degree

$2 ,$ the degree of the remainder (poly.) sought for, must be

less than $2.$

Therefore, we suppose that, the remainder is $a x + b .$

Now, if $q \left(x\right)$ is the quotient poly., then, by the Remainder Theorem,

we have, $p \left(x\right) = \left(x - 1\right) \left(x + 2\right) q \left(x\right) + \left(a x + b\right) , \mathmr{and} ,$

$3 {x}^{5} - 5 {x}^{2} + 4 x + 1 = \left(x - 1\right) \left(x + 2\right) q \left(x\right) + \left(a x + b\right) \ldots \ldots \left(\star\right) .$

$\left(\star\right) \text{ holds good } \forall x \in \mathbb{R} .$

We prefer, x=1, and, x=-2!

Sub.ing, $x = 1$ in $\left(\star\right) , 3 - 5 + 4 + 1 = 0 + \left(a + b\right) , \mathmr{and} ,$

$a + b = 3. \ldots \ldots \ldots \ldots \ldots \ldots \left({\star}_{1}\right) .$

Similarly, sub.inf $x = - 2$ in $p \left(x\right)$ gives,

$2 a - b = 123. \ldots \ldots \ldots \ldots \ldots \left({\star}_{2}\right) .$

Solving $\left({\star}_{1}\right) \mathmr{and} \left({\star}_{2}\right) \text{ for } a \mathmr{and} b ,$ we get,

$a = 42 \mathmr{and} b = - 39.$

These give us the desired remainder,

$42 x - 39 = 3 \left(14 x - 13\right) .$

Enjoy Maths.!