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 $x + 3.$

Pushy Jul 15, 2019

#1**+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)

CPhill Jul 15, 2019

#2**+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...

tommarvoloriddle Jul 15, 2019