If ax³ + bx³ +x - 6 has (x+2) as factor and leaves remainder 4 when divided by x-2. Find a and b.

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Solution

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)

Adding equations (1) and (2),
⇒ 4a=0
⇒ a=0
Therefore ,b = 2

Then, equation will be 2x² + x - 6

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