Question

The polynomial x^2 -4x +3 is a factor of
x^2 +(a-4)x^2 + (3-4a)x +3
calculate the value of a constant a

Answer 1

hmm i guess 1 is a zeros of the first quadratic, so it must be a zero of the second one as well.
replace x by 1, set the result equal to zero, and solve for a

Answer 2

i have an answer i think is right. what do you get?

Answer 3

i dont get it hahah

Answer 4

oh ok

Answer 5

\[x^2 +(a-4)x^2 + (3-4a)x +3 =( x^2 -4x +3)(\text{ something})\]

Answer 6

if you let
\[x=1\] you get on the right hand side
\[(1-4+3)(\text{something})=0\]

Answer 7

which means (since they are equal) the left hand side must be zero as well if you let x = 1

Answer 8

so replace x on the left hand side by 1, set the result equal zero, and you can solve for a

Answer 9

you with me so far?

Answer 10

ahhhhh ok ! yes am with you!

Answer 11

is it just sorta like trial and era?

Answer 12

ok so now we replace x by one and get
\[1+(a-4)+ (3-4a) +3 =0\]
\[3-3a=0\]
\[a=1\]

Answer 13

no not really
if we know that 1 is a zero of
\[x^2-4x+3\] then since this is a factor of the other polynomial, it must be a zero of that one as well

Answer 14

ahhhh ok!! :D

Answer 15

and could it be anything else?

Answer 16

like x=3

Answer 17

notice that if we had just plugged in any number we would not be in such good shape. it is the zero that makes this work.

Answer 18

maybe. lets try it

Answer 19

\[x^2-4x+3=(x-3)(x-1)\] so 3 is a zero as well. so we can replace x by 3 on the left hand side and see what we get
\[3^2+(z-4)\times 3^2+(3-4a)\times 3+3=0\]

Answer 20

hmm looks like i get
\[a=-5\]

Answer 21

man i must have messed this up, or else there is no solution. in fact is
\[x^2-4x+3\] is a factor, then both 1 and 3 must give zero. but you get different values for a depending on whether you use 1 or 3, so i am going to say there is no solution to this

Answer 22

you sure it is
\[x^2 +(a-4)x^2 + (3-4a)x +3 \]

Answer 23

yapp am sure it is that!!

Answer 24

hmm but then in the first one shoudnt it be plus 4 in the end and not 3?
so its 1+(a-4) + (3-4a) +4?

Answer 25

ok so i am thinking there is no solution to this one
if
\[x^2 +(a-4)x^2 + (3-4a)x +3 =(x-1)(x-3)(\text{something})\] then that means they are equal for all values of x.
now if x = 1 i get a =1 and if x = 3 i get z = -5, and since they are not the same, there is a big problem here

Answer 26

no, if x = 1 you should get
\[1+(a-4)+(3-4a)+3=0\]

Answer 27

ohh ok hahaha

Answer 28

hmm then i guess there is a huge problem

Answer 29

i have another problem as well. the constant on both sides is 3

Answer 30

really?? :O

Answer 31

oh my this question is bad retrice

Answer 32

and both sides are of degree 2
so if the left hand side is a multiple of the right hand side, then it is a constant multiple

Answer 33

yes i see your point

Answer 34

and since they both have constant 3, that constant is 1

Answer 35

in other words if one is a multiple of the other, they are equal

Answer 36

YAH YAH

Answer 37

is this question from a book of some kind?

Answer 38

nope my teacher uploaded it! it says exercises from may examinations 2004 haha

Answer 39

i got a whole pellet to do.. :O plus questions in the book

Answer 40

maybe there is a typo

Answer 41

because look, if the sides are equal, that means
\[x^2 +(a-4)x^2 + (3-4a)x +3=x^2-4x+3\] which means
\[a-4=0\] or
\[a=4\] but if
\[a=4\]then
\[3-4a=-13\] which is impossible since it has to be -4

what class is this? i really think there is a mistake here
can you upload the question directly?

Answer 42

if you send me the link i will look more carefully, because i would hate to think that i am telling you something that is wrong. but at this point i would say there is a mistake

Answer 43

ok its not a link though its a pdf file..

Answer 44

ok attach it and i will take a look