Question

Can someone help me solve these questions?

a) If (x-2) is a factor of 2x3-3x+k, find k.

b) When ax3-4x2+5x-3 is divided by (x+2) and (x-1), the remainders are equal. Find a.

Answer 1

if x-2 is a factor of 2x3-3x+k,
then
2(2)^3 -3(2) +k =0

Answer 2

use if F(x) is divided by x-a then the Remainder is given by F(a)
in case if x-a is a factor of F(x) then F(a) =0

Answer 3

Can you please help me solve it mat?

Answer 4

@rcatabijan, Did you find the value of k yet?

Answer 5

Not yet, I need help in doing so..

Answer 6

is it because you moved "k"? to the other side?

Answer 7

by shifting it to the other side right? would the answer be k = -2(2)^3 + 3^2? we can say we found k?

Answer 8

can you please help me simplify it

Answer 9

k = 16 + 9
k = 25

Answer 10

yeah?

Answer 11

oh yes because of the negative.. so it would be -7

Answer 12

how about for b), how would I go about starting the question?

Answer 13

Mind staying with (a) for a few minutes longer? Have you checked to determine whether x-2 is truly a factor of 2x^3-3x+k when k=-7?

Answer 14

But, my Hero, wouldn't -3(2)=-6, not -9?

Answer 15

Ugh, stems from a mistake I made earlier.

Answer 16

k = -2(2)^3 + 3(2)
k = -2(8) + 6
k = -16 + 6
k = -10

Answer 17

I found that synthetic div. works wonderfully well in solving this problem. Having solved it, I noticed that the previous posters came up with k=-7, which is incorrect. k=-10 is right.

Answer 18

What's wrong with @matricked method?

Answer 19

By the way, you never got back to me regarding the file I sent you earlier I notice.

Answer 20

Have you read part B, and if so, have you begun solving it? I'm curious how you'd proceed. Again I'm starting out with synth. div. because this method is usually so short and efficient.
I did look at that file and will look at it again. I believe I found one small error similar to your -3(2)=-9.

Answer 21

I know. I wrote that mistake while trying to avoid it. I know exactly what you're talking about.

Answer 22

@mathmale, if you keep referring back to that mistake so much, I'm may feel the urge to delete your comment. Constantly referring back to that mistake isn't helping anyone as much as it is insulting.

Answer 23

You acknowledged the mistake, now move on.

Answer 24

Why not correct the doc and then share it with me again? Another thought I had while reading it was that some might be more willing to read it were you to provide some verbal explanations and directions instead of just the math symbolism.
Frankly, I'm taken aback at your outburst of anger. I had had NO intention of dwelling on it. I'd written the above suggestion when your angry message came through. Perhaps this is not the right venue to be talking about your method.

Answer 25

I never said I was angry about anything. I'll get back to you later though.

Answer 26

(Somewhat at a loss for words.) All right.