# If P(x)=2x^3-3x^2-5x+6 is divided by x-2, then (1) what is the remainder; (2) what is the quotient and (3) how we can express P(x) in terms of factors?

Mar 27, 2017

See below.

#### Explanation:

I think some sign or coefficient are not well fitted because

assuming $q \left(x\right) = a {x}^{2} + b x + c$ and substituting

$2 {x}^{3} - 3 {x}^{2} - 5 x + 6 = \left(a {x}^{2} + b x + c\right) \left(x - 2\right) + r$ or

$2 {x}^{3} - 3 {x}^{2} - 5 x + 6 = a {x}^{3} + \left(b - 2 a\right) {x}^{2} + \left(c - 2 b\right) x + r - 2 c$

Identifying coefficients,

$\left\{\begin{matrix}a = 2 \\ b - 2 a = - 3 \\ c - 2 b = 5 \\ r - 2 c = 6\end{matrix}\right.$

Solving for $a , b , c , r$ we get

$a = 2 , b = 1 , c = - 3 , r = 0$ so

$q \left(x\right) = 2 {x}^{2} + x - 3$ and $r = 0$

Solving now $q \left(x\right) = 0$ we obtain

$x = \left(- \frac{3}{2} , 1\right)$ or

$q \left(x\right) = 2 \left(x + \frac{3}{2}\right) \left(x - 1\right)$

Concerning $p \left(x\right) = 2 {x}^{3} - 3 {x}^{2} - 5 x + 6 = 0$

we know that

$p \left(x\right) = q \left(x\right) \left(x - 2\right)$

or

$p \left(x\right) = 2 \left(x + \frac{3}{2}\right) \left(x - 1\right) \left(x - 2\right)$

Mar 28, 2017
1. $r = 0$, $P \left(x\right) = 2 {x}^{3} - 3 {x}^{2} - 5 x + 6$
2. $q \left(x\right) = \left(2 x + 3\right) \left(x - 1\right)$
3. $P \left(x\right) = \left(2 x + 3\right) \left(x - 1\right) \left(x - 2\right)$

#### Explanation:

$P \left(x\right) = 2 {x}^{3} - 3 {x}^{2} - 5 x + 6 = q \left(x\right) \left(x - 2\right) + r$

Using remainder theorem $P \left(2\right) = r$

and therefore $r = 2 \cdot {2}^{3} - 3 \cdot {2}^{2} - 5 \cdot 2 + 6$

$= 16 - 12 - 10 + 6 = 0$ and

$P \left(x\right) = 2 {x}^{3} - 3 {x}^{2} - 5 x + 6 = q \left(x\right) \left(x - 2\right) + 0$

or $q \left(x\right) \left(x - 2\right) = 2 {x}^{3} - 3 {x}^{2} - 5 x + 6$

$= 2 {x}^{3} - 3 {x}^{2} - 5 x + 6$

$= 2 {x}^{2} \left(x - 2\right) + x \left(x - 2\right) - 3 \left(x - 2\right)$

$= \left(2 {x}^{2} + x - 3\right) \left(x - 2\right)$

$= \left(2 {x}^{2} + 3 x - 2 x - 3\right) \left(x - 2\right)$

$= \left(x \left(2 x + 3\right) - 1 \left(2 x + 3\right)\right) \left(x - 2\right)$

$= \left(2 x + 3\right) \left(x - 1\right) \left(x - 2\right)$

and $P \left(x\right) = \left(2 x + 3\right) \left(x - 1\right) \left(x - 2\right)$