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When $p(x) = Ax^5 + Bx^3 + Cx + 4$ is divided by $x - 3,$ the remainder is 11. Find the remainder when $p(x)$ is divided by $x + 3.$

 Jul 15, 2019
 #1
avatar+103999 
+2

 

 

Dividing p(x) by x - 3 implies that

 

p(3)  =   Remainder  ....so....

 

A(3)^5  + B(3)^3 + C(3) + 4  =  11  ⇒   243A + 27B +3C + 4  = 11 ⇒ [ 243A + 27B + 3C]  =  7

 

And p(-3)  = Remainder

 

A(-3)^5  + B(-3)^3 + C(-3) + 4  = Remainder   ⇒  -243A - 27B - 3C  =  Remainder - 4  ⇒

 

- [ 243A - 27B + 3C]  = Remainder - 4

 

- [ 7 ]    = Remainder - 4      add 4 to both sides

 

-3  = Remainder =  p(-3) 

 

 

cool cool cool

 Jul 15, 2019
 #2
avatar+1338 
+5

Fewer dollar signs version of the problem:

 

When \(p(x) = Ax^5 + Bx^3 + Cx + 4\) is divided by \(x - 3,\) the remainder is 11. Find the remainder when \(p(x)\) is divided by \(x+3\).

 

 

The dollar signs are ok for now, but when you get to like a huge problem with a fraction polynomial, we really can't understand it... That would not be the most ideal form of the problem...

That may result in a wrong answer...

 Jul 15, 2019

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