Help with Remainder and Factor Theorem?
Need help with this Remainder and Factor Theorem question.
When a polynomial P(x) is divided by (x-2) and (x+3), the remainders are 4 and -26 respectively. Find the remainder when P(x) is divided by (x-2)(x+3). Please show steps, thanks.
- jsjsLv 59 years agoFavourite answer
The given information says that there exists polynomials Q(x) and R(x) such that
P(x) = (x-2) Q(x) + 4,
P(x) = (x+3) R(x) - 26.
Plugging x=-3 into the second equation, we get P(-3)=-26. Plugging into the first, we get
-26 = P(-3) = (-5) Q(-3) + 4.
Solving for Q(-3), we have Q(-3)=6. Therefore 6 is the remainder when Q(x) is divided by (x+3). In other words, there exists a polynomial S(x) such that
Q(x) = (x+3) S(x) + 6.
Plugging this in above, we have
P(x)= (x-2) [(x+3)S(x) + 6] + 4 = (x-2)(x+3) S(x) + 6(x-2) + 4 = (x-2)(x+3) S(x) + 6x - 8.
Therefore the remainder when P(x) is divided by (x-2)(x+3) is 6x - 8.
- 4 years ago
I divided via (x - 2) and have been given a the remainder of 2a + b which I positioned equivalent to one million So 2a + b = one million and b = one million-2a I divided via (x+one million) and have been given a the remainder of b - (a+3) which i positioned equivalent to twenty-eight So b - a - 3 = 28 positioned b = one million -2a provides one million - 2a - a - 3 = 28 So 3a = - 30 so a = -10 replace for a provides -20 + b = one million So b = 21 replace those values in unique equation x^3 -2x^3 -10x +21 and divide via (x -3 ) provides (x-3)(x^2 +x -7) devoid of the rest This exhibits (x-3) is a element