VCE Methods Question Thread!

[gumby97]

Hi guys

Find the value of a for which (1-2a)x^2 + 5ax + (a-1)(a-8) is divisible by (x-2) but not by (x-1)

help would be much appreciated, never come across a question like this before

since it's divisible by x-2
solve for P(2) = 0
should get a = 3 or 4 but becareful, it said that this equation cannot be divisable by x-1 so
solve for P(1) =/ 0
(a-3)^2=/0, a =/3
so since a=/3, a=4 which is the other option

Thanks man

Just two more if you could please

a. prove that the expression x^3 + (k-1)x^2 + (k-9)x -7 is divisible b x+1 for all values of k

                                     and
b. the expression 4x^3 +ax^2 -5x +b leaves remainders of -8 and 10 when divided by (2x-3) and (x-3) respectively.  Calculate the the values of a and b

[aznxD]

a) Let p(x) = x^3 + (k-1)x^2 + (k-9)x -7
p(-1) = 0
Therefore p(x) is divisible by x+1 for all values of k

b) Let p(x) = 4x^3 +ax^2 -5x +b
p(3/2) = -8
Therefore 9a/4 + b +6 = -8  (1)
p(3) = 10
9a + b + 93 = 10   (2)

Solve (1) and (2) simultaneously and you get:
a = -140/9
b = 39

[butene]

a) Let p(x) = x^3 + (k-1)x^2 + (k-9)x -7
p(-1) = 0
Therefore p(x) is divisible by x+1 for all values of k

b) Let p(x) = 4x^3 +ax^2 -5x +b
p(3/2) = -8
Therefore 9a/4 + b +6 = -8  (1)
p(3) = 10
9a + b + 93 = 10   (2)

Solve (1) and (2) simultaneously and you get:
a = -140/9
b = 39

wut, i got a=-92/9 and b=9 unless i did some careless mistakes o_O

[dc302]

a) Let p(x) = x^3 + (k-1)x^2 + (k-9)x -7
p(-1) = 0
Therefore p(x) is divisible by x+1 for all values of k

b) Let p(x) = 4x^3 +ax^2 -5x +b
p(3/2) = -8
Therefore 9a/4 + b +6 = -8  (1)
p(3) = 10
9a + b + 93 = 10   (2)

Solve (1) and (2) simultaneously and you get:
a = -140/9
b = 39

wut, i got a=-92/9 and b=9 unless i did some careless mistakes o_O

No you're right.

[aznxD]

a) Let p(x) = x^3 + (k-1)x^2 + (k-9)x -7
p(-1) = 0
Therefore p(x) is divisible by x+1 for all values of k

b) Let p(x) = 4x^3 +ax^2 -5x +b
p(3/2) = -8
Therefore 9a/4 + b +6 = -8  (1)
p(3) = 10
9a + b + 93 = 10   (2)

Solve (1) and (2) simultaneously and you get:
a = -140/9
b = 39

wut, i got a=-92/9 and b=9 unless i did some careless mistakes o_O

Whoops, sorry my silly mistake >.<
I made 9a/4 + b +6 = 10 and 9a + b + 93 = -8
Instead of 9a/4 + b +6 = -8 and 9a + b + 93 = 10

[Bhootnike]

Thanks man

Just two more if you could please

a. prove that the expression x^3 + (k-1)x^2 + (k-9)x -7 is divisible b x+1 for all values of k

                                   

an easier method discussed earlier to this same question, or similar , is that:

use remainder theorem:

f(x) =
you're told to use x + 1, so thats f(-1)

therefore:

f(-1) = -1 + (k-1)-1^2 + (k-9)(-1) -7
        = -1 + k - 1 -k +9 - 7
        = 0

therefore, since 0 remainder, it is divisble for values for k.

edit. just saw that theyve used the same method.. hehe woops

[Bhootnike]

b. the expression 4x^3 +ax^2 -5x +b leaves remainders of -8 and 10 when divided by (2x-3) and (x-3) respectively.  Calculate the the values of a and b

id just use synthetic division

        |  4     a    -5           b
 3/2  |        6   1.5a +9    6 + 2.25a
        |________________________
          4     a+6   4 +1.5a   b+ 6+ 2.25a

therefore

b+6 + 2.25a = -8

b + 2.25a = -14

similarly, you now divide by 3:

        |  4     a         -5           b
 3     |         12      3a+36     9a +93
        |________________________
          4     a+12   3a +31   b +9a +93

therefore: b + 9a + 93 = 10

b+ 9a = -83       ---------2

therefore, b = -83 -9a ----3

now sub 3 into 1.


b + 2.25a = -14
-83 -9a + 2.25a = -14
-6.75a = 69

a = -10.22222222222222

therefore, b = -83 - 9(-10.2222222222222)
b= 9



sorry ive used decimals. just cause im typing haha.. cbs using fractions.
but answers are equivalent to 62/9 and yeah

just my approach.
quick and easy..

[paulsterio]

I'd say just use the remainder theorem!




calculate P(3/2) = ...
let that = -8 (remainder)


calculate P(3) = ...
let that = 10 (remainder)


you'll have 2 equations with 2 unknowns, CAS it or solve it by hand

[Phy124]

id just use synthetic division

I'm not sure you're allowed to use this method in Methods? edit: disregard this sentence, I just read another post on the forum and apparently it is allowed.

Remainder theorem is easy enough and I'd say that would be the method VCAA would be looking for, if they gave a similar question.

[aznxD]

The remainder theorem is definitely much more efficient in terms of these type of questions.

[Bhootnike]

id just use synthetic division

I'm not sure you're allowed to use this method in Methods? edit: disregard this sentence, I just read another post on the forum and apparently it is allowed.

Remainder theorem is easy enough and I'd say that would be the method VCAA would be looking for, if they gave a similar question.

Nah its not allowed according to pi !
but i wouldnt assume such questions to be in MC ?
in S.A though, i'd definitely use remainder theorem.. ,
i just like using the synthetic div. since my school allows it :p

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[pi]

Lol synthetic division is NOT allowed

[Hutchoo]

Lol synthetic division is NOT allowed

Long division isn't that much harder/tedious than synthetic division anyway.

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[dc302]

Lol synthetic division is NOT allowed

VCAA has a document that says synthetic is allowed...just found out from the other thread.

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[aznxD]

Lol synthetic division is NOT allowed

Long division isn't that much harder/tedious than synthetic division anyway.

^ This.

Moderator action: removed real name, sorry for the inconvenience