polynomials (1 Viewer)

[shkspeare]

When a polynomial P(x) is divided by x-@, the remainder is a² and when divided by x-b, the remainder is b². Show that when P(x) is divided by (x-@)(x-b), the remainder is (@+b)x-@b

 

[Affinity]

first

P(x) = (x-a)(x-b)Q(x) + (cx+d) for some cx+d

then:

P(a) = a^2

P(b)= b^2

by the remainder theorem.

so if you
ca+d=a^2
cb +d = b^2

subtract the bottom equation from the one above

c(a-b) = a^2 - b^2
c= a+b

solve for d by substituting in ca + d = a^2

a^2 + ab + d = a^2
d = - ab

the remainder is

cx+d = (a+b)x -ab

second question: don't think so

 

[shkspeare]

thx!!!

mm accidentalyl deleted teh question

it was "if odd polynomials (not DEGREE) and even polynomials have a root, "a" another root mst be -a"

(unless its a = 0)

 

[KeypadSDM]

For both odd and even:
f(a) = 0

Even:

f(a) = f(-a)
f(-a) = 0

Odd:
f(a) = -f(-a)
-f(-a) = 0
f(-a) = 0