If ax³+bx²+x-6 has (x+2) as a …
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Gaurav Seth 4 years, 3 months ago
Let p(x) = ax³ + bx² + x - 6
Now,
(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)
Now, 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
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