Question

If x^2 - 3x + 2 divides x^3 - 6x^2 + ax + b exactly , then find the value of 'a' and 'b'

Answer 1

Answer :

Given :

$\longrightarrow{Quadratic \:expression = {x}^{2} - 3x + 2}$

$\longrightarrow{Cubic \:expression = {x}^{3} - 6 {x}^{2} + ax + b }$

Required to find :

$\boxed{Values \:of\: "a" \:and\: "b"}$

Explanation :

In the given with 2 polynomial expression ;

1. Quadratic expression .

2. Cubic expression .

Here,

We have to factorise the Quadratic expression in order to find its factor ,

Using this factors we can find the value of a and b

By, substituting their respective values in the Quadratic expression .

Hence, we can solve this question .

Solution :

Given :

$\longrightarrow{Quadratic\: expression = {x}^{2} - 3x + 2 }$

Here, we have to factorise this by splitting the middle term .

The middle term should be splited in such a manner that their product should be equal to the last term and first term.

Similarly, their sum should be equal to the middle term .

Hence,

$\longrightarrow{{x}^{2} - 3x + 2}$

$\longrightarrow{{x}^{2} - 2x - 1x + 2 }$

$\longrightarrow{x (x - 2) - 1 ( x - 2) }$

$\implies{\boxed{(x - 2)}{\boxed{(x - 1)}}}$

Therefore, the factors of the cubic expression are (x - 2) & (x - 1) .

Hence,

The given cubic expression is ;

$\longrightarrow{p(x) = {x}^{3} - {6x}^{2} + ax + b }$

Now, let's consider (x - 2) as the factor

Let , x - 2 = 0

x = 2

Hence,

P(2) =

$\longrightarrow{{(2)}^{3} - 6{(2)}^2 + a (2) + b = 0 }$

$\longrightarrow{8 - 6(4) + 2a + b = 0 }$

$\longrightarrow{8 - 24 + 2a + b = 0 }$

$\longrightarrow{- 16 + 2a + b = 0 }$

$\longrightarrow{ b = - 2a + 16 }{\rightarrow{equation -- 1}}$

Similarly,

Now, let's consider ( x - 1 ) as the factor .

Let, x - 1 = 0

x = 1

Hence,

P(1) =

$\longrightarrow{{(1)}^{3} - 6{(1)}^2 + a (1) + b = 0 }$

$\longrightarrow{1 - 6(1) + a + b = 0 }$

$\longrightarrow{1 - 6 + a + b = 0 }$

$\longrightarrow{ - 5 + a + b = 0 }$

$\longrightarrow{ - 5 + a - 2a + 16 = 0 \:\:\:\:\:\:\:\:\Rightarrow From equation 1}$

$\longrightarrow{ - 1a + 11 = 0 }$

$\longrightarrow{ - 1a = - 11 }$

$\longrightarrow{ \cancel {-}a = \cancel {-}11}$

$\longrightarrow{\textsf{Negative signs get cancelled on both sides}}$

Hence,

$\longrightarrow{a = 11}$

similarly,

Value of b is,

$\Rightarrow{-2a + 16 }$

$\Rightarrow{-2(11) + 16 }$

$\Rightarrow{-22 + 16}$

$\Rightarrow{ - 6 }$

Value of b is -6

Therefore,

$\longrightarrow{\boxed{\therefore{Value\: of\: a \:is \:11}}}$

$\longrightarrow{\boxed{\therefore{Value\: of\: b \:is \:-\:6}}}$