Find $p$ if the remainder when $(x^2 + px + 13)$ divided by $(x-2)$ is twice the remainder when it is divided...
$begingroup$
As far as I understand, this should be a simultaneous equation, although I'm not really sure where to go beyond that. I made the remainder of $f(-2)=r$ and $f(2)=2r$, and substituted the respective values.
$x^2+px+13$
With $(x-2)$
$f(2)=(2)^2+p(2)+13$
$f(2)=17+2p=2r$
With $(x+2)$
$f(-2)=(-2)^2+p(-2)+13$
$f(-2)=17-2p=r$
I guess I could do a simultaneous here, and I got $17/6$ as my result, but I'm not sure if that's correct.
proof-verification polynomials
edited Dec 17 '18 at 16:08
Tianlalu
3,08421138
asked Dec 17 '18 at 16:04
Roo23Roo23
213
$endgroup$
add a comment |
$begingroup$
As far as I understand, this should be a simultaneous equation, although I'm not really sure where to go beyond that. I made the remainder of $f(-2)=r$ and $f(2)=2r$, and substituted the respective values.
$x^2+px+13$
With $(x-2)$
$f(2)=(2)^2+p(2)+13$
$f(2)=17+2p=2r$
With $(x+2)$
$f(-2)=(-2)^2+p(-2)+13$
$f(-2)=17-2p=r$
I guess I could do a simultaneous here, and I got $17/6$ as my result, but I'm not sure if that's correct.
proof-verification polynomials
edited Dec 17 '18 at 16:08
Tianlalu
3,08421138
asked Dec 17 '18 at 16:04
Roo23Roo23
213
$endgroup$
1
$begingroup$
looks good to me
$endgroup$
– randomgirl
Dec 17 '18 at 16:20
add a comment |
$begingroup$
As far as I understand, this should be a simultaneous equation, although I'm not really sure where to go beyond that. I made the remainder of $f(-2)=r$ and $f(2)=2r$, and substituted the respective values.
$x^2+px+13$
With $(x-2)$
$f(2)=(2)^2+p(2)+13$
$f(2)=17+2p=2r$
With $(x+2)$
$f(-2)=(-2)^2+p(-2)+13$
$f(-2)=17-2p=r$
I guess I could do a simultaneous here, and I got $17/6$ as my result, but I'm not sure if that's correct.
proof-verification polynomials
edited Dec 17 '18 at 16:08
Tianlalu
3,08421138
asked Dec 17 '18 at 16:04
Roo23Roo23
213
$endgroup$
As far as I understand, this should be a simultaneous equation, although I'm not really sure where to go beyond that. I made the remainder of $f(-2)=r$ and $f(2)=2r$, and substituted the respective values.
$x^2+px+13$
With $(x-2)$
$f(2)=(2)^2+p(2)+13$
$f(2)=17+2p=2r$
With $(x+2)$
$f(-2)=(-2)^2+p(-2)+13$
$f(-2)=17-2p=r$
I guess I could do a simultaneous here, and I got $17/6$ as my result, but I'm not sure if that's correct.
proof-verification polynomials
proof-verification polynomials
edited Dec 17 '18 at 16:08
Tianlalu
3,08421138
asked Dec 17 '18 at 16:04
Roo23Roo23
213
edited Dec 17 '18 at 16:08
Tianlalu
3,08421138
asked Dec 17 '18 at 16:04
Roo23Roo23
213
edited Dec 17 '18 at 16:08
Tianlalu
3,08421138
edited Dec 17 '18 at 16:08
Tianlalu
3,08421138
edited Dec 17 '18 at 16:08
Tianlalu
3,08421138
3,08421138
asked Dec 17 '18 at 16:04
Roo23Roo23
213
asked Dec 17 '18 at 16:04
Roo23Roo23
213
asked Dec 17 '18 at 16:04
Roo23Roo23
213
213
1
$begingroup$
looks good to me
$endgroup$
– randomgirl
Dec 17 '18 at 16:20
add a comment |
1
$begingroup$
looks good to me
$endgroup$
– randomgirl
Dec 17 '18 at 16:20
1
1
$begingroup$
looks good to me
$endgroup$
– randomgirl
Dec 17 '18 at 16:20
$begingroup$
looks good to me
$endgroup$
– randomgirl
Dec 17 '18 at 16:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are making it too hard.
$f(2) = 17+2p,
f(-2) = 17-2p$.
So you want
$17+2p = 2(17-2p)
=34-4p$
so
$17=6p$
or
$p = 17/6$
as you got.
answered Dec 17 '18 at 18:15
marty cohenmarty cohen
74.3k549128
$endgroup$
add a comment |
$begingroup$
HINT:
$$x^2+px+13=(x+p+2)(x-2)+(17+2p)=(x+p-2)(x-2)+(17-2p)$$
Via polynomial long division.
answered Dec 17 '18 at 16:21
Rhys HughesRhys Hughes
6,9801630
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are making it too hard.
$f(2) = 17+2p,
f(-2) = 17-2p$.
So you want
$17+2p = 2(17-2p)
=34-4p$
so
$17=6p$
or
$p = 17/6$
as you got.
answered Dec 17 '18 at 18:15
marty cohenmarty cohen
74.3k549128
$endgroup$
add a comment |
$begingroup$
You are making it too hard.
$f(2) = 17+2p,
f(-2) = 17-2p$.
So you want
$17+2p = 2(17-2p)
=34-4p$
so
$17=6p$
or
$p = 17/6$
as you got.
answered Dec 17 '18 at 18:15
marty cohenmarty cohen
74.3k549128
$endgroup$
add a comment |
$begingroup$
You are making it too hard.
$f(2) = 17+2p,
f(-2) = 17-2p$.
So you want
$17+2p = 2(17-2p)
=34-4p$
so
$17=6p$
or
$p = 17/6$
as you got.
answered Dec 17 '18 at 18:15
marty cohenmarty cohen
74.3k549128
$endgroup$
You are making it too hard.
$f(2) = 17+2p,
f(-2) = 17-2p$.
So you want
$17+2p = 2(17-2p)
=34-4p$
so
$17=6p$
or
$p = 17/6$
as you got.
answered Dec 17 '18 at 18:15
marty cohenmarty cohen
74.3k549128
answered Dec 17 '18 at 18:15
marty cohenmarty cohen
74.3k549128
answered Dec 17 '18 at 18:15
marty cohenmarty cohen
74.3k549128
answered Dec 17 '18 at 18:15
marty cohenmarty cohen
74.3k549128
74.3k549128
add a comment |
add a comment |
$begingroup$
HINT:
$$x^2+px+13=(x+p+2)(x-2)+(17+2p)=(x+p-2)(x-2)+(17-2p)$$
Via polynomial long division.
answered Dec 17 '18 at 16:21
Rhys HughesRhys Hughes
6,9801630
$endgroup$
add a comment |
$begingroup$
HINT:
$$x^2+px+13=(x+p+2)(x-2)+(17+2p)=(x+p-2)(x-2)+(17-2p)$$
Via polynomial long division.
answered Dec 17 '18 at 16:21
Rhys HughesRhys Hughes
6,9801630
$endgroup$
add a comment |
$begingroup$
HINT:
$$x^2+px+13=(x+p+2)(x-2)+(17+2p)=(x+p-2)(x-2)+(17-2p)$$
Via polynomial long division.
answered Dec 17 '18 at 16:21
Rhys HughesRhys Hughes
6,9801630
$endgroup$
HINT:
$$x^2+px+13=(x+p+2)(x-2)+(17+2p)=(x+p-2)(x-2)+(17-2p)$$
Via polynomial long division.
answered Dec 17 '18 at 16:21
Rhys HughesRhys Hughes
6,9801630
answered Dec 17 '18 at 16:21
Rhys HughesRhys Hughes
6,9801630
answered Dec 17 '18 at 16:21
Rhys HughesRhys Hughes
6,9801630
answered Dec 17 '18 at 16:21
Rhys HughesRhys Hughes
6,9801630
6,9801630
add a comment |
add a comment |
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1
$begingroup$
looks good to me
$endgroup$
– randomgirl
Dec 17 '18 at 16:20