# Math Help - help solve 2

1. ## help solve 2

The polynomial P(x)=x2+ax +b has a zero at x =2. When P(x)is divided by x +1, the remainder is 18.Find the values of a and b.

2. Originally Posted by andy69
The polynomial P(x)=x2+ax +b has a zero at x =2. When P(x)is divided by x +1, the remainder is 18.Find the values of a and b.

You have $P(x) = x^2 + ax + b$.

By the Remainder and Factor Theorems:

If there is a root at $x = 2$, then $P(2) = 0$.

If, when you divide by $x + 1$ you get a remainder of $18$, then $P(-1) = 18$.

So you have:

$2^2 + 2a + b = 0$

$(-1)^2 - a + b = 18$.

Simplify and solve these equations simultaneously for $a$ and $b$.

3. Originally Posted by Prove It
You have $P(x) = x^2 + ax + b$.

By the Remainder and Factor Theorems:

If there is a root at $x = 2$, then $P(2) = 0$.

If, when you divide by $x + 1$ you get a remainder of $18$, then $P(-1) = 18$.

So you have:

$2^2 + 2a + b = 0$

$(-1)^2 - a + b = 18$.

Simplify and solve these equations simultaneously for $a$ and $b$.
so i get 4+2a+b=0(1) and 1-a+b=18(2)
then i make 1-a+b=18(times by 2) which is 2-2a+2b=36(3)
by elimination method
(3)+(1)
6+3b=36
3b=30
b=10
sub b=10 into (1)
4+2a+10=0
2a=-14
a=-7

is this right?

4. Originally Posted by andy69
so i get 4+2a+b=0(1) and 1-a+b=18(2)
then i make 1-a+b=18(times by 2) which is 2-2a+2b=36(3)
by elimination method
(3)+(1)
6+3b=36
3b=30
b=10
sub b=10 into (1)
4+2a+10=0
2a=-14
a=-7

is this right?
Yes, well done.

5. Originally Posted by Prove It
Yes, well done.
thanks for the help