Question

Let f(x)=ax^3+bx^2+cx+d and suppose that r is a root of the equation f(x)=0. Show that r-h is a root of f(x+h)=0.

Answer 1

we have below by factor/remainder theorem :
`if f(k) = 0, then "k" is a root`

say,
g(x) = f(x+h) = a(x+h)^3+b(x+h)^2+c(x+h)+d
g(r-h) = ?

Answer 2

g(r-h)=f(r)... Thus, it equals 0??

Answer 3

@rational ?

Answer 4

Exactly ! thats the proof

Answer 5

@rational Thanks and.... what is the difference between the factor and remainder theorem if you mind answering?

Answer 6

thats a very good question !

Answer 7

remainder theorem speaks first

Answer 8

it talks about the remainder in long division of a "polynomian" by a "linear polynomial" :

\[\large \dfrac{P(x)}{x-a}\]

Answer 9

the remainder of above long division is same as P(a)

Answer 10

thats remainder theorem ^^

Answer 11

is that clear ?

Answer 12

Yes, I think so (:
Thank you

Answer 13

we're not done with factor theorem yet

Answer 14

@rational It's okay, I read the definition in the textbook and I remembered it! (: Summer makes me forget a lot ;o Also, will you be willing to maybe solve the last two parts of this problem?

Answer 15

good :)

factor theorem says this :

if \(\large (x-r)\) is a factor of \(\large P(x)\), then \(\large P(r) = 0\)

and

if \(\large P(r) = 0\), then \(\large (x-r)\) is a factor of \(\large P(x)\)

Answer 16

yeah wil try ask

Answer 17

Thank you (: Show that -r is a root of the equation f(-x)=0

Answer 18

use the same trick

Answer 19

say,
g(x) = f(-x) = a(-x)^3+b(-x)^2+c(-x)+d
g(-r) = ?

Answer 20

say,
g(x) = f(-x) = a(-x)^3+b(-x)^2+c(-x)+d
g(-r) = f(--r) = a(--r)^3+b(--r)^2+c(--r)+d
= ar^3+br^2+cr+d
= 0

Answer 21

Show that kr is a root of the equation f(x/k)=0

Answer 22

Oh and thank you (:

Answer 23

@rational I got it. Thanks