If (ax3+bx2+x-6) has (x+2) as a factor and leaves remainder 4, when divided by (x-2), find the values of a and b.

It is given that

(x + 2)

is the factor of

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

and leaves remainder 4, when divided by

(x - 2)

.
Let,

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

.
Find the value of x which satisfies the given polynomial.
Since

(x + 2)

is the factor of a given polynomial, it must satisfy the given polynomial.
⇒x+2=0
⇒x=-2
At x = -2, we get,
⇒

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

⇒

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

- 4 (2 a - b + 2) = 0

2 a - b = - 2

…...eq(i)
Given that

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

leaves remainder 4, when divided by

(x - 2)

.
Therefore, by putting

x = 2

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

,leaves the remainder 4. Hence,

p (2) = 4

.
⇒

p (2) = a {(2)}^{3} + b {(2)}^{2} + (2) - 6 = 4

⇒

8 a + 4 b - 4 = 4

⇒

8 a + 4 b = 8

⇒

2 a + b = 2

……eq(ii)
Solving eq(i) and eq(ii),
⇒eq(i)+eq(ii), we get,
⇒

(2 a - b

(2 a + b) = - 2 + 2

⇒

4 a = 0

⇒

a = 0

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

2 (0) + b = 2

⇒

b = 2

Therefore, option 1 is correct.

If $(a x^{3} + b x^{2} + x - 6)$ has $(x + 2)$ as a factor and leaves remainder 4, when divided by $(x - 2)$ , find the values of a and b.