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

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Answer 1

Answer:

3 and 4

step-by-step explanation:

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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