When the polynomial in p(x) is divided by (x + a), the remainder equals p(a)

This is a false statement:

Step-by-step explanation:

According to Remainder Theorem dividing the polynomial by some linear factor x + a, where a is just some number. As a result of the long polynomial division, you end up with some polynomial answer q(x) (the "q" standing for "the quotient polynomial") and some polynomial remainder r(x).

P(x)= (x+/-a) q(x)+r(x)

P(x)=(x+a) q(x)+r(x). Note that for x=-a

P(-a)=(-a+a) q(-a)+r(-a)= 0* q(-a)+ r(-a)

P(-a)=r(-a)

It means that P(-a) is the remainder not P(a)

Thus the given statement is false

x

step-by-step explanation:

hello from mrbilldoesmath!

answer:

the domain is the set { -2, 0, 2, 3}

discussion:

the domain is the set of x values in the ordered pairs (x,y). taking the x, or first component, of each ordered pair gives   -2, 0, 2, 3.

you,

mrb