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# Polynomial remaindery stuff.

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538
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idk if the upload of the picture worked (if it didnt someone help me understand how to use it)

Find k if the remainder is 3 when 5x^2-10x+k is divided by x-1

I have several questions like this one so im more looking for an explanation than a solution :) thank you!

Jun 1, 2015

#2
+27374
+10

If 5x2 - 10x + k has remainder 3 when divided by x - 1 then we must have:

5x2 - 10x + k = (x - 1)*m + 3  where m is an integer.

When x = 1 the right-hand side is just 3 and m disappears so put x = 1 in the left-hand side as well to get

5 - 10 + k = 3

or k = 8

(Chris beat me to it!)

Incidentally, you can't upload images as long as you stay Anonymous.  Why not register, then you will have access to the upload images function.

.

Jun 1, 2015

#1
+94548
+10

Notice that if we substituted 1 into the polynomial, we could find the value of k that would make 1 a root....in other words, we could find the k value such that, P(1) = 0....... (or, put another way, would make (x-1) a divisor of the polynomial with no remainder).......

So.....using the same idea.......we want to find the value of k that makes P(1)  = 3

So P(1)  =  5(1)^2 - 10(1) + k = 3   →  5 - 10 + k = 3  →  -5 + k = 3 →  k = 8

Test this for yourself using synthetic division.......you will find that you get a remainder of 3 when k = 8 and you divide by 1.....

Jun 1, 2015
#2
+27374
+10

If 5x2 - 10x + k has remainder 3 when divided by x - 1 then we must have:

5x2 - 10x + k = (x - 1)*m + 3  where m is an integer.

When x = 1 the right-hand side is just 3 and m disappears so put x = 1 in the left-hand side as well to get

5 - 10 + k = 3

or k = 8

(Chris beat me to it!)

Incidentally, you can't upload images as long as you stay Anonymous.  Why not register, then you will have access to the upload images function.

.

Alan Jun 1, 2015
#3
+94548
0

That's OK, Alan.....you gave me 3 points....I'll give you 3......!!!

Jun 1, 2015