Question

The polynomial ax^3 + bx^2 + x-6 when divided by (x+2) and (x-2) leaves remainder 0 and 4 respectively find value of a and b
It is very urgent plz answer

Answer 1

Answer:

x + 2 = 0.

x = - 2

p(x) = ax^3 + bx^2 + x - 6

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

0 = -8 a + 4b - 8

-2a + b - 2 =0

-2a + b = 2

b = 2 + 2a ----(1)

x - 2 = 0

x = 2

p(x) = ax^3 + bx^2 + x - 6

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

4 = 8a + 4b - 4

2a + b - 1 = 1

2a + b = 2

b = 2 - 2a ----(2)

substituting eq (1) in eq (2)

2 + 2a = 2 - 2a

4a = 0

a = 0 ---(3)

substituting eq (3) in eq (2)

b = 2 - 2(0)

b = 2

a = 0 and b = 2

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Answer 2

Step-by-step explanation:

Let p(x) = ax³ + bx² + x - 6

A/C to question,

(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0

So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0

⇒-8a + 4b - 8 = 0

⇒ 2a - b + 2 = 0 -------------(1)

again, question said that if we p(x) is divided by ( x -2) then it leaves remainder 4.

so, P(2) = a(2)³ + b(2)² + 2 - 6 = 4

⇒8a + 4b - 4 = 4

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

solve equations (1) and (2),

4a = 0 ⇒a = 0 and b = 2

Then, equation will be 2x² + x - 6

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