What is the remainder when x³ + ax² + 2x + 9 is divided by x + 1

Whenx³ + ax² + 2x + 9 is divided by x - 1 the remainder is 7.
What is the remainder when x³ + ax² + 2x + 9 is divided by x + 1?

So I know we add the opposite I thought, to get P(-1) = (-1)³ +a(-1)²+2(-1)+9
But I get a+6 as my answer and don't think this is solving remainder but rather a....
How do i solve the remainder?

Could someone show me the steps...
I've covered the whole polynomial chapter today, and am loving it and having "fun" if you can call it that for math, but can't get this problem now!

By the Remainder Theorem, the remainder from dividing p(x) = x³ + ax² + 2x + 9 by x - 1 is given by p(1) = 1 + a + 2 + 9 = a + 12.

However, we are told that this remainder equals 7.
Hence, a + 12 = 7 ==> a = -5.

Thus, p(x) = x³ - 5x² + 2x + 9.

So, using the Remainder Theorem again, the remainder of p(x) by (x + 1) is
p(-1) = -1 - 5 - 2 + 9 = 1.

I hope this helps!

You're doing it right!
If you look back at the first one, you would work out the remainder as 12 +a, but since you know that the actual remainder is 7, 12 + a= 7 => a = -5

you've then done the next steps to find out the remainder is a + 6, a = -5 and voila the remainder is equal to 1

Hope this helped! :)

f(x)=x³ + ax² + 2x + 9
f(x)=q(x)(x-1)+7
hence f(1)=1³ + a*1² + 2*1 + 9=q(1)(1-1)+7=q(1)*0+7=7
1³ + a*1² + 2*1 + 9=7
a=-5

f(x)=x³ - 5x² + 2x + 9
f(x)=g(x)(x+1)+r
f(-1)=(-1)³ - 5(-1)² + 2(-1) + 9 =1

Use synthetic division

–1 | ......1......a......2......9
...................–1...1–a...a–3
..........----------------------------…
............1....a–1...3–a...a+6
a + 6 is the correct remainder

P(1) = 7
1 + a + 2 +9 = 7
a = -5

P(-1) = R
R = -1 -5(1) + 2(-1) + 9
= -1 +5 -2 +9
=1