If x3+ax2+bx+6 has (x-2) as a factor and leaves a remainder 3 when divided by(x-3), then find the values of a and b.

It is given that

(x - 2)

is the factor of

(x^{3} + a x^{2} + bx + 6)

and leaves remainder 4, when divided by

(x - 3)

.
Let,

p (x) = x^{3} + a x^{2} + bx + 6

.
Finding the value of x which satisfies the given polynomial,
Since

(x - 2)

is the factor of a given polynomial, therefore, it must satisfy the given polynomial.
At x = 2, we get,
⇒

p (2) = {(2)}^{3} + a {(2)}^{2} + b (2) + 6 = 0

⇒

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

⇒

p (2) = 2 (2 a + b + 7) = 0

⇒

(2 a + b + 7) = 0

…...eq(i)
We know that

a x^{3} + b x^{2} + x - 6

leaves remainder 3, when divided by

(x + 2)

.
Therefore, putting

x = - 2

a x^{3} + b x^{2} + x - 6

leaves the remainder 3. Hence,

p (- 2) = 3

.
⇒

p (- 2) = {(- 2)}^{3} + a {(- 2)}^{2} + b (- 2) + 6 = 3

⇒

27 + 9 a + 3 b + 6 - 3 = 0

\Rightarrow (3 a + b + 10) = 0

……eq(ii)
Therefore, solving eq(i) and eq(ii),
⇒eq(ii)-eq(i), we get,
⇒

(3 a + b + 10) - (2 a + b + 7) = 0

⇒

a + 0 + 3 = 0

⇒

a = - 3

Putting value of ‘a’ in eq(ii), we get,
⇒

(3 (- 3) + b + 10) = 0

⇒

b = - 1

Hence, the value of

a = - 3, b = - 1

.
Therefore, option 1 is correct.

If $(x^{3} + a x^{2} + bx + 6)$ has $(x - 2)$ as a factor and leaves a remainder 3 when divided by $(x - 3)$ , then find the values of $a$ and $b$ .