# help solve 2

#### 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.

#### Prove It

MHF Helper
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 $$\displaystyle P(x) = x^2 + ax + b$$.

By the Remainder and Factor Theorems:

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

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

So you have:

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

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

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

#### andy69

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

By the Remainder and Factor Theorems:

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

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

So you have:

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

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

Simplify and solve these equations simultaneously for $$\displaystyle a$$ and $$\displaystyle b$$.
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?

#### Prove It

MHF Helper
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. (Clapping)

• andy69