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The expression x^3 + ax^2 + bx + 3 is exactly divisible by x+3
but it leaves a remainder of 91 when divided by x-4.
What is the remainder when it is divided by x+2?
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The solution is to apply the Remainder theorem several times.
(1) Due to the Remainder theorem, first condition means that the numner -3 is the root of the given polynomial.
It gives you this equation
(-3)^3 + a*(-3)^2 + b*(-3) + 3 = 0,
which is equivalent to
9a - 3b = 24, or 3a - b = 8. (1)
(2) Due to the same theorem, second condition means that the value of the given polynomial is 91 at x= 4.
It gives you this equation
4^3 + a*4^2 + b*4 + 3 = 91,
which is equivalent to
16a + 4b = 24, or 4a + b = 6. (2)
(3) Adding equations(1) and (2), we get 7a = 14, a = 14/7 = 2.
Then from (1), 3*2 - b = 8, b = 6 - 8 = -2.
So, the polynomial is p(x) = x^3 + 2*x^2 - 2*x + 3.
(4) Due to the Remainder theorem, the remainder of the polynomial when it is divided by x+2 is the value of the polynomial at x= -2, i.e.
(-2)^3 + 2*(-2)^2 -2*(-2) + 3 = 7. ANSWER
Solved.
