+96 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 $x + 3.$
+129852
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)
+1713 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...
