Question

The expression ax^3-8x^2+bx+6 is divisible by x^2-2x-3. Find values of a & b.
@ikram002p

Answer 1

so (ax^3-8x^2+bx+6)= (ax-M) (x^2-2x-3)
a.t M is devides 6 so it might be 1,2,3,6,-1,-2,-3,-6

Answer 2

so
\((ax-M) (x^2-2x-3)=ax^3-2ax^2-3ax-Mx^3+2Mx+3M\)

Answer 3

\(ax^3-8x^2+bx+6 =ax^3-(M+2a)x^2+(2M-3a)x +3M\)

Answer 4

3M=6

so what is M value ?

Answer 5

2

Answer 6

yes , so ur getting the qn ?

Answer 7

2M-3a=b
M+2a=8

Answer 8

M actually refers to...?

Answer 9

its a variable i added it so if
\( ax^3-8x^2+bx+6 \) is divisible by \( x^2-2x-3 \)
then \( ax^3-8x^2+bx+6= (\text{other term} )(x^2-2x-3)\)
ok

Answer 10

(so other ) term can be
(ax-M)
we dnt know M value yet

Answer 11

so the equation will be like this
\( ax^3-8x^2+bx+6 =(ax-M)( x^2-2x-3)\)

Answer 12

expand
\((ax-M)( x^2-2x-3)=?\)

Answer 13

Understood
So, this's the method to solve this type of question..

Answer 14

yes ^^

Answer 15

Another question...

Answer 16

wait but did u find a,b for this ?

Answer 17

sure ask ^^

Answer 18

when polynomial x^2+ax+b is divided by (x-1) & (x+2), its remainder is 4 n 5 respectively.
Find the values of a n b

Answer 19

so \(x^2+ax+b =(x-M)+\large \frac {4}{x-1}\)

so \(x^2+ax+b =(x-N)+ \large \frac {5}{x-2}\)

Answer 20

its like 9/2
u can write the answer like 4+ 1/2
since

Answer 21

use long Division

Answer 22

Okay, I'll try to solve this question using this method..

Answer 23

:D
do u have other method ?

Answer 24

Nope

Answer 25

if yes i would be happy to learn it :P

Answer 26

ok solve this , and tell me when u get it ^_^

Answer 27

sry i made a typo brb

Answer 28

alternative method for the first problem,
x^2-2x-3 =0
gives x= 3,-1

so, plug in x=3 in ax^3-8x^2+bx+6 =0
get one equation in a and b

plug in x=-1 in ax^3-8x^2+bx+6 =0
get another equation in a and b

solve simultaneously.

Answer 29

This method seems to be much simpler

Answer 30

alternative method for
"when polynomial x^2+ax+b is divided by (x-1) & (x+2), its remainder is 4 n 5 respectively.
Find the values of a n b"

plug in x=1,
in x^2+ax+b =4
get one equation in a and b

plug in x=-2,
in x^2+ax+b =5
get another equation in a and b

solve simultaneously

Answer 31

i have used remainder theorem .

Answer 32

i made a typo , i ment to write equation like this

\(\large \frac{ x^2+ax+b}{x-1} =(x-M)+\large \frac {4}{x-1}\)

\(\large \frac{ x^2+ax+b}{x-2} =(x-MN)+\large \frac {5}{x-2}\)

Answer 33

\(\large \frac{ x^2+ax+b}{x-2} =(x- N)+\large \frac {5}{x-2}\) *

Answer 34

okay

Answer 35

yeah so first one it will be
\(x^2+ax+b =(x-M ) (x-1) + 4\)
\(x^2+ax+b =(x-N ) (x+2) + 5\)

Answer 36

^_^

Answer 37

for the 2nd question.. a=2/3, b= 7/3
Another question:
The expression x^3-4x^2+x+6 & x^3-3x^2+2x+k have common factor.Find possible values of k.

Answer 38

Can I use remainder theorem to solve this question
@hartnn

Answer 39

too much work for the first question

Answer 40

u can do it in 3 steps probably

Answer 41

you can use remainder theorem there, but it won't help you find k

Answer 42

is it mono factor ?

Answer 43

maybe

Answer 44

heard about rational root theorem ?

Answer 45

yeah , that will work

Answer 46

No.

Answer 47

then which topic does this question belong to ?

Answer 48

Chapter: Functions
Sub topic is factor theorem..
I don't mind learning this rational root....

Answer 49

x^3-4x^2+x+6 =(x-a)(....)
x^3-3x^2+2x+k =(x-a)(...)
let (x-a ) be the common factor , then a should be a factor of both 6,k

Answer 50

factorize 6:-
then
possible values for a is
\(\pm1,\pm 2 , \pm 3\)

Answer 51

but
k= n a
mmmm we need to find a upper bound for k

Answer 52

would u plz factorize this equation ?
x^3-4x^2+x+6

Answer 53

what i can think of as an alternative way is

since we take 'x-a' as a factor,
we can plug in x=a in both expressions and equate them

a^3-4a^2+a+6 = a^3-3a^2+2a+k

this gives you quadratic in a and k

Answer 54

and since roots must be real
you can do
b^2-4ac >0 to find range of k

Answer 55

but ultimately possible values of k will be from factors of 6 only.

Answer 56

\( x^3-4x^2+x+6 =(x+1)(x-2)(x-3)\)
so three cases :-
\( x^3-3x^2+2x+k =(x+1)(...)\)
\( x^3-3x^2+2x+k =(x-2)(...)\)
\( x^3-3x^2+2x+k =(x-3)(...)\)

Answer 57

what must I do nxt

Answer 58

next make usfull of long Division

Answer 59

Okay..will try now