Question

6. (x-2) is a factor of the expression x? + ax? + bx + 6. When the expression is divided by (x-3), it leaves the remainder 3. Find the values of a and b. ​

Answer 1

Correct Question :-

(x-2) is a factor of the expression x³ + ax² + bx + 6. When the expression is divided by (x-3), it leaves the remainder 3.

To Find :-

Values of a and b

Solution :-

(x - 2) = 0

x = 2

Putting x = 2

(2)³ + a(2)² + b(2) + 6 = 0

8 + a(4) + 2b + 6 = 0

8 + 4a + 2b + 6 = 0

14 + 4a + 2b = 0

Dividing by 2

14 + 4a + 2b/2 = 0/2

7 + 2a + b = 0

2a + b = -7 (i)

When divided by (x - 3) the remainder is 3

(x - 3) = 0

x = 3

(3)³ + a(3)² + b(3) + 6 = 3

27 + a(9) + 3b + 6 = 0

33 + 9a + 3b = 0

Dividing by 3

33 + 9a + 3b/3 = 3/3

11 + 3a + b = 1

3a + b = 1 - 11

3a + b = -10 (ii)

On subtracting 1 and 2

2a + b - 3a - b = -7 - (-10)

2a - 3a = -7 + 10

-a = 3

a = -3

Using 2

3a + b = -10

3(-3) + b = -10

-9 + b = -10

b = -10 + 9

b = -1

Value of a and b are -3 and -1 respectively

Answer 2

Correct Question :-

(x-2) is a factor of the expression x³ + ax² + bx + 6. When the expression is divided by (x-3), it leaves the remainder 3. Find the values of a and b.

Required Answer :-

Now it is given that (x-2) is a factor of the expression x³ + ax² + bx + 6

${\sf{:\implies {x} - 2}}$

${\sf{:\implies {x} = 2}}$

And ,

${\sf{:\implies f( x) \: =x^{3} + ax ^{2} + bx + 6 }}$

★ ( f(2)=0 )

by putting the value of x = 2

${\sf{:\implies 2^{3} + a(2) ^{2} + b(2) + 6 }} = 0$

${\sf{:\implies8 + 4a + 2b + 6 }} = 0$

${\sf{:\implies 4a + 2b = - 14}}$

${\sf{:\implies 2(2a + b) = -14}}$

${\sf{:\implies2a + b = -7}}$ (equation 1)

It is given that when expression is divided by (x-3), it leaves the remainder 3

So ,

${\sf{:\implies {x} - 3}}$

${\sf{:\implies {x} = 3}}$

and ,

${\sf{:\implies f( x) \: =x^{3} + ax ^{2} + bx + 6 }}$

★ ( f(3) = 3 )

by putting the value of x = 3

${\sf{:\implies 3^{3} + a(3) ^{2} + b(3)+ 6 = 3 }}$

${\sf{:\implies 27 + 9a + 3b+ 6 = 3 }}$

${\sf{:\implies 9a + 3b = - 30 }}$

${\sf{:\implies 3(3a + b )= - 30 }}$

${\sf{:\implies 3a + b = - 10 }}$ (equation 2)

On subtraction Equation 1 from equation 2 we get ,

⠀⠀⠀2a + b = -7

⠀–⠀3a + b = -10

⠀-a = 3 or a = -3

Now , we know that the value of a = -3 . So , from equation 1 we can find the value of b.

${\sf{:\implies 2a + b = - 7}} ... \sf(a = - 3)$

${\sf{:\implies - 6 + b = - 7}}$

${\sf{:\implies b = - 1}}$

Hence ,

• a = -3
• b = -1