# maths question - will give rep..when i can

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#1
p(x) = (x+2)^5

Find the remainder when p(x) is divided by (x+3)

how do i do this w/o finding the expansion of (x+2)^5.

thanks.
0
16 years ago
#2
(Original post by ShOcKzZ)
p(x) = (x+2)^5

Find the remainder when p(x) is divided by (x+3)

how do i do this w/o finding the expansion of (x+2)^5.

thanks.
Let p(-3), and then you have (-3+2)^5, Remainder = -1
0
16 years ago
#3
Are you familiar with

p(x)/(x-a) = q(x) + R/(x-a)

Ie when you divide a polynomial by a linear factor x-a you get a polynomial plus some remainder over x-a.

This => p(x) = q(x)(x-a) + R
If we take x=a .......p(a) = 0 + R

So the remainder of p divided by x-a is equal to p(a).

This should be enough to answer the question
0
#4
(Original post by 2776)
Let p(-3), and then you have (-3+2)^5, Remainder = -1
thats what i thought. but, if you expand (x+2)^5, you get:

p(x) = x^5 + 10x^4 + 40x^3 + 80X^2 + 80x + 25
so:

p(-3) = -8
0
#5
(Original post by It'sPhil...)
Are you familiar with

p(x)/(x-a) = q(x) + R/(x-a)

Ie when you divide a polynomial by a linear factor x-a you get a polynomial plus some remainder over x-a.

This => p(x) = q(x)(x-a) + R
If we take x=a .......p(a) = 0 + R

So the remainder of p divided by x-a is equal to p(a).

This should be enough to answer the question
no, i didn't know that, lemme try it..
0
16 years ago
#6
(Original post by It'sPhil...)
So the remainder of p divided by x-a is equal to p(a).

This should be enough to answer the question
Since (x-a) = [x+(-a)]

Therefore, [x+(-3)], and so remainder = a, therefore (-3) is correct.
0
16 years ago
#7
(Original post by ShOcKzZ)
thats what i thought. but, if you expand (x+2)^5, you get:

p(x) = x^5 + 10x^4 + 40x^3 + 80X^2 + 80x + 25
so:

p(-3) = -8
2^5 usually equals 32 not 25
0
#8
(Original post by It'sPhil...)
2^5 usually equals 32 not 25
oh ****

lol..so the -1 was correct..
0
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