Question

PLEASE HELP
Given a, b, and c are integers such that 0 < a < b, and given that the polynomial
p(x)=x(x – a)(x – b) – 17 is divisible by (x – c), find the value of a + b +c.

Answer 1

Have you considered multiplying (x-a)(x-b)(x-c)?

Answer 2

yes, but i dont know how it helps or relates

Answer 3

p(x) is divisible by (x-c) gives you \[c(c-a)(c-b) = 17\]

Answer 4

wait why is it that you can switch out c for x?

Answer 5

by factor theorem, if (x-c) is a factor of p(x) then p(c) = 0

Answer 6

ok i will look that up. is the answer straightforward after this?
\[c (c ^{2}+ab-ac-bc) = c ^{3}+abc-c ^{2}a-c ^{2}b = 17\]

Answer 7

expanding makes it tedious, consider using the fact that 17 is a prime instead

Answer 8

the only factors of 17 are \(\pm 1, \pm 17\)

so the left hand side factors must be one of these

Answer 9

should i just try out numbers or is there a better way?

Answer 10

ofcourse there will be a better way, lets think...

Answer 11

well, there are four cases correct?
Given \[c(c-a)(c-b) = 17\]
c = 17, (c-a) = 1, (c-b) = 1
c = 17, (c-a) = -1, (c-b) = -1
c = -17, (c-a) = -1, (c-b) = 1
c = -17, (c-a) = 1, (c-b) = -1

Answer 12

they are impossible because they give you a=b but you're given a < b

Answer 13

oh right, 0<a<b

Answer 14

so only case 3 works?

Answer 15

so looks like \(c\) has to be +1

Answer 16

how did you get that?

Answer 17

you have already worked that c cannot be +-17, yes ?

Answer 18

sorry, i dont understand why that is so.

Answer 19

c = 17, (c-a) = 1, (c-b) = 1
c = 17, (c-a) = -1, (c-b) = -1

these two give you "c-a = c-b"
cancelling c both sides you get a = b so reject

Answer 20

ok, but why cant c = -17?

Answer 21

since a,b are positive, when c is negative, both c-a and c-b are also negative yes ?

Answer 22

Omg i completely disregarded the 0 in the inequality. Is that the same reason for which
c=-1 is not viable?

Answer 23

Yep! c cannot be negative because that forces c-a and c-b to be negative.

Answer 24

And the product of 3 negative numbers is negative but you want a positive number on right hand side

Answer 25

so c must equal 1?

Answer 26

Yes, next find a, b

Answer 27

and (c-a) = 17 or 1 and (c-b) = 1 or 17

Answer 28

waiiiit dooesnt that force either a or b to be negative?

Answer 29

c = 1 forces both (c-a) and (c-b) to be negative

Answer 30

ok so a, b = 2 or 18 but since a<b, a=2 and b=18?

Answer 31

is that the only possible solution?

Answer 32

Looks good!

Answer 33

you can make the solution short and neat by eliminating "c" being negative in the start itself with a simple argument

Answer 34

Would that argument be that if c is negative, either (a-b) or (a-c) must also be negative, forcing either a or b to be negative which contradicts 0<a<b?

Answer 35

i mean (c-a) or (c-b) in the above comment

Answer 36

you have c(c-a)(c-b) = 17.

if c is negative, both (c-a) and (c-b) must also be negative, forcing the left hand side to be negative. so reject.

Answer 37

oh i get it. Thanks so much i appreciate it!!!!!! Could you help me with one more problem i have?

Answer 38

I'll try, ask..

Answer 39

The odd numbers from 1 to 17 can be used to build a 3 x 3 magic square (a square in which the rows and columns have the same sum). If the 1, 5, and 13 are located as shown, what is the value of x?

Answer 40

i tried filling the voids with variables but i couldnt get anything solvable

Answer 41

can you use the numbers more than once ?

Answer 42

no the numbers are limited to 1,3,5,7,9,11,13,15,17 used only once each

Answer 43

Okay first observation is the sum has to be >= 20

Answer 44

i believe that the bottom middle spot must be 17, because when i let the middle spot be equal to c, 5+c+13 = 1+c+17 is that correct?

Answer 45

Oh you can use only odd numbers, wait i was trying all numbers

Answer 46

only the numbers 1,3,5,7,9,11,13,15,17

Answer 47

i think this is valid: