Question

For a polynomial p(x), the value of p(3) is -2. Which of the following must be true about p(x)?

Answer 1

A) x-5 is a factor of p(x)
B) x-2 is a factor of p(x)
C) x+2 is a factor of p(x)
D) The remainder when p(x) is divided by x-3 is -2.

Answer 2

Remainder Theorem:

if p(x) is divided by x-k, and the remainder is r, then p(k) = r

For example, if p(10) = 9 then p(x) divided by x-10 leads to a remainder of 9

Answer 3

If the remainder is 0, then x-k is a factor of p(x)

Answer 4

Not enough information is provided on whether x=5, x=2 or x=-2 are roots of p(x). I am left with D but I don't fully get why is it true. P(x)/x-k=q(x)+r Where q(x) is the quotient and r is the remainder, but can't the remainder be a polynomial as well.

Answer 5

`but can't the remainder be a polynomial as well.`
any single number without variables attached is technically a polynomial. It's a polynomial of degree 0. So the remainder is always a polynomial

Answer 6

p(x)/(x-k) = q(x) + r/(x-k)

If you multiply both sides by (x-k) and simplify, you'll get

p(x) = (x-k)*q(x) + r

Answer 7

now plug in x = k

p(x) = (x-k)*q(x) + r
p(k) = (k-k)*q(k) + r ... replace every x with k
p(k) = (0)*q(k) + r
p(k) = 0 + r
p(k) = r

Answer 8

What I have meant was that r could be a polynomial of a degree of more than zero. Why is it r/x-k?

Answer 9

it's r/(x-k) to indicate a fractional portion.

For instance

10/7 = 1 + 3/7

the '1' is the quotient q(x)
the '3' is the remainder r(x)

Answer 10

we can multiply both sides of `10/7 = 1 + 3/7` by 7 to get

10/7 = 1 + 3/7
7*(10/7) = 7*(1 + 3/7)
10 = 7*(1) + 7*(3/7)
10 = 7*(1) + 3


The "7" plays the role as the "x-k"
The "1" is the quotient
The "3" is the remainder

Answer 11

the remainder r could be some higher degree polynomial, but only if you're dividing polynomials like

(x^3+6x+5) over (x-3)

notice we have a cubic over a first degree polynomial. The degree of the remainder could be as high as 2

Answer 12

Got it, thanks @jim_thompson5910

Answer 13

no problem