When x^3+kx^2+9x+3 is divided by x−2 the remainder is 5. Use the remainder theorem to find the value of k.

When x^3+kx^2+9x+3 is divided by x−2 the remainder is 5. Use the remainder theorem to find the value of k.

Firstly, please explain wth the remainder thereom is,

thank you

If you are dividing by x - 2 then x must = 2

Use x = 2 in the polynomial
x^3+kx^2+9x+3 and put it = 5

2^3 + (k x 2^2) + 9 x 2 + 3 = 5

8 + 4k + 18 + 3 = 5
29 + 4k = 5
4k = 5 - 29
4k = -24
k = -6

remainder theory is the nice one lol, use the information given as substitution to work out as proof or missing values

Remainder theorem: f(x) = q(x)g(x) + r(x)

g(x) is the divisor, q(x) the quotient and r(x) the remainder. Naturally r(x) is of lesser degree than g(x). In the case of linear g(x) then r(x) is some constant r.
What this boils down to is that if f(x) is divided by (x-a) then f(a) = r

Hence you know that f(2) = 5

divide through by x-2 carry the k through and you will get a remainder of 4k+29 we know that is = 5
set equal and solve
4k+29=5
k = -6 plug in and check
i did this using http://www.wolframalpha.com/input/?i=%28…

f(x) = x^3 + kx^2 + 9x + 3

By remainder theorem, f(2) = 5:
f(x) / (x-2) = g(x) + 5/(x - 2)
f(x) = (x - 2) g(x) + 5
f(2) = 0 * g(x) + 5
f(2) = 5

8 + 4k + 18 + 3 = 5
4k + 29 = 5
4k = -24
k = -6

Mαthmφm

a/b = c
then
c * b = a

(x - 2) * (ax^2 + bx + c + 5/(x - 2)) = x^3 + kx^2 + 9x + 3
(x - 2) * (ax^2 + bx + c) + 5 = x^3 + kx^2 + 9x + 3
(x - 2) * (ax^2 + bx + c) = x^3 + kx^2 + 9x - 2
ax^3 + bx^2 + cx - 2ax^2 - 2bx - 2c = x^3 + kx^2 + 9x - 2
x^3 * (a) + x^2 * (b - 2a) + x * (c - 2b) - 2c = x^3 + kx^2 + 9x - 2

ax^3 = 1x^3
a = 1

x^2 * (b - 2a) = kx^2
b - 2a = k
b - 2 * 1 = k
b - 2 = k

x * (c - 2b) = 9x
c - 2b = 9

-2c = -2
c = 1

c - 2b = 9
1 - 2b = 9
-8 = 2b
b = -4

b - 2 = k
-4 - 2 = k
-6 = k

Maybe do your own homework like everyone else, if you are having trouble ask your teacher not us