# Question help (1 Viewer)

#### Dashdorm24

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

Would appreciate all help, thanks

#### CM_Tutor

##### Moderator
Moderator
If $\bg_white P(x)$ has a zero at $\bg_white x = 2$ then $\bg_white (x - 2)$ must be a factor of $\bg_white P(x)$. We know that $\bg_white P(x)$ is monic (coefficient of $\bg_white x^2$ is 1) and so it can be factorised as $\bg_white P(x) = (x - 2)(x - k)$ for some constant $\bg_white k$.

Now, when $\bg_white P(x)$ is divided by $\bg_white (x + 1)$, the remainder is 18, meaning that

\bg_white \begin{align*} P(-1) &= 18 \\ (-1 - 2)(-1 - k) &= 18 \\ -3 \times -(k + 1) &= 18 \\ k + 1 &= 6 \\ k &= 5 \end{align*}

So, our polynomial is

$\bg_white P(x) = (x - 2)(x - 5) = x^2 - 7x + 10$

and so $\bg_white a = -7$ and $\bg_white b = 10$

#### davidgoes4wce

##### Well-Known Member
The polynomial P(x) = x^2 + ax + b has a zero at x=2. When P(x) is divided by x+1, the remainder is 18. Find the value of a and b.

Would appreciate all help, thanks
I had a slightly different way of doing it

P(-1)=18
P(2)=0

18=(-1)^2+a(-1)+b
18=1-a+b
17=-a+b equation A

0=2^2+2a+b
-4=2a+b equation B

subtract equation A -B

21=-3a
a=-7

sub a=-7, 17=-(-7)+b
b=24

#### CM_Tutor

##### Moderator
Moderator
I had a slightly different way of doing it

P(-1)=18
P(2)=0

18=(-1)^2+a(-1)+b
18=1-a+b
17=-a+b equation A

0=2^2+2a+b
-4=2a+b equation B

subtract equation A -B

21=-3a
a=-7

sub a=-7, 17=-(-7)+b
b=24
Yes, this is the more obvious way... my approach was more suited to the question where P(x) is a polynomial of degree $\bg_white n \geqslant 2$.

But, there is an error in the last step of your calculation. It should say:

\bg_white \begin{align*} 17 &= -(-7) + b \\ 17 &= 7 + b \\ 17 - 7 &= b \\ b &= 10 \end{align*}