View Full Version : Remainder Theorem
danishiya
Oct 10, 2009, 07:28 PM
for what value of b will be polynomial p(x) = -2x^3 + bx^2 - 5x + 2 have the same remainder when it is divided by x- 2 and by x+1 ?
galactus
Oct 11, 2009, 03:24 AM
Divide it by x-2 and get:
-2x^{2}+bx-4x+2b-13+\underbrace{\frac{4b-24}{x-2}}_{\text{remainder}}
Divide by x+1:
-2x^{2}+bx+2x-b-7+\underbrace{\frac{b+9}{x+1}}_{\text{remainder}}
Can you finish?
Look at the remainder terms. Equate the numerators in the remainder terms and solve for b.
danishiya
Oct 11, 2009, 11:33 AM
noo. Don't you have to sub 1 into 2 to get the other value.
galactus
Oct 11, 2009, 12:23 PM
No.
4b-24=b+9
Unknown008
Oct 14, 2009, 11:04 AM
If you use the remainder theorem, then you apply it. It says that for a given polynomial f(x), the remainder, when f(x) is divided by (x - a), is given by f(a).
Hence;
p(x) = -2x^3 + bx^2 - 5x + 2
Substituting 2;
p(2) = -2(2)^3 + b(2)^2 - 5(2) + 2 = -16 + 4b - 10 + 2 = 4b - 24
Substituting -1;
p(-1) = -2(-1)^3 + b(-1)^2 - 5(-1) + 2 = 2 + b + 5 + 2 = 9 + b
If both remainder are equal, then:
4b - 24= 9 + b
Solve for b.