Question

if x³ - 6x² + ax + b is divisible by (x² - 3x + 2), then the values of a and b are:
1. a = -6 and b = -11
2. a = -11 and b = 6
3. a= 11 and b = -6
4. a = 6 and b = 11

Answer 1

Correct Answer - Option 3 : a= 11 and b = -6

Given:

Dividend = x3 - 6x2 + ax + b

Divisor = x2 - 3x + 2

Concept used:

Remainder theorem: It says if a polynomial is divided by (x – a), then remainder will be after substituting the value of [(x - a) = 0 ⇒ x = a] in given polynomial.

If a polynomial is dived by (x + a), then remainder will be after substituting the value of (x + a = 0 ⇒ x = - a) in given polynomial.

If a polynomial is divided by (ax + b), then remainder will be after substituting the value of [ax + b = 0 ⇒ ax = - b ⇒ x = (-b/a)] in given polynomial.

If a polynomial is divided by (ax - b), then remainder will be after substituting the value of [ax - b = 0 ⇒ ax = b ⇒ x = (b/a)] in given polynomial.

Calculation:

Divisor = x2 - 3x + 2

Now we calculate the factor of x2 - 3x + 2

Then, x2 - 3x + 2 = 0

⇒ x2 - 2x - x + 2 = 0

⇒ x(x - 2) - 1(x - 2) = 0

⇒ (x - 2)(x - 1) = 0

Then value of x will be (x - 2) = 0

⇒ x = 2

And the value of x will be also (x - 1) = 0

⇒ x = 1

Now the value of x is 2 and 1.

After putting of the value of x = 2 in the given dividend we get,

x3 - 6x2 + ax + b = 0

⇒ 2³ - 6 × 2² + a × 2 + b = 0

⇒ 8 - 24 + 2a + b = 0

⇒ 2a + b = 16      ----(i)

After putting of the value of x = 1 in the given dividend we get,

x3 - 6x2 + ax + b = 0

⇒ 13 - 6 × 12 + a × 1 + b = 0

⇒ 1 - 6 + a + b = 0

⇒ a + b = 5      ----(ii)

After subtracting of equation (ii) from equation (i), we get

2a - a = 16 - 5

⇒ a = 11

Using the value of a = 11 in equation (ii) we get,

11 + b = 5

⇒ b = 5 - 11

⇒ b = - 6

∴ The value of a and b are 11 and - 6 respectively.

Comments

Equation 1 is 2a+b=16
And, Equation 2 is a+b=5,
While, subtracting Equation 2 from Equation 1, why you should take equation 1 = 2a, and equation 2 = a,
Where is b??