# Question 33e19

Mar 17, 2017

The divisor is

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

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

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

$= \left(x - 3\right) \left(x - 2\right)$
so it has two linear factors $\left(x - 3\right) \mathmr{and} \left(x - 2\right)$

The polynomial of $x$ to be divided is

$P \left(x\right) = a {x}^{3} - 9 {x}^{2} + b x + 3 a$
This is to be exactly divisible by the divisor containing two linear factors $\left(x - 3\right) \mathmr{and} \left(x - 2\right)$.

So

$P \left(3\right) = 0$

$\implies a \cdot {3}^{3} - 9 \cdot {3}^{2} + b \cdot 3 + 3 a = 0$

$\implies 30 a + 3 b = 729$

$\implies 10 a + b = 243. \ldots \ldots . \left(1\right)$

Again

$P \left(2\right) = 0$

$\implies a \cdot {2}^{3} - 9 \cdot {2}^{2} + b \cdot 2 + 3 a = 0$

$\implies 11 a + 2 b = 36. \ldots \ldots . \left(2\right)$

Multiplying (1) by 2 and subtracting (2) from the product we get

$20 a - 11 a = 486 - 36$

$\implies 9 a = 450$

$\implies a = \frac{450}{9} = 50$

Inserting the value of a in equation (1)

=>10xx50+b=243)#

$\implies b = 243 - 500 = - 257$

Mar 17, 2017

$a = 2 , b = 7$

#### Explanation:

If $a {x}^{3} - 9 {x}^{2} + b x + 3 a$ is exactly divisible by ${x}^{2} - 5 x + 6$ then

$a {x}^{3} - 9 {x}^{2} + b x + 3 a \equiv 0 \mod {x}^{2} - 5 x + 6$

then $\exists \left(c x + d\right) | a {x}^{3} - 9 {x}^{2} + b x + 3 a = \left(c x + d\right) \left({x}^{2} - 5 x + 6\right)$

or

$\left(a - c\right) {x}^{3} + \left(5 c - 9 - d\right) {x}^{2} + \left(5 d - 6 c + b\right) x + 3 a - 6 d = 0$

Solving

$\left\{\begin{matrix}a - c = 0 \\ 5 c - 9 - d = 0 \\ 5 d - 6 c + b = 0 \\ 3 a - 6 d = 0\end{matrix}\right.$

we have

$a = 2 , b = 7 , c = 2 , d = 1$