Question

question

Answer 1

\(\large \color{black}{\begin{align} & \normalsize \text{if}\ x^2+ax+b\ \text{leaves same remainder } 5\hspace{.33em}\\~\\
& \normalsize \text{when divided by}\ (x-1)\ \text{or}\ (x+1)\hspace{.33em}\\~\\
& \normalsize \text{then the values of}\ a\ \text{and}\ b\ \text{are}\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

\[f(x)=x^2+ax+b \\ f(1)=1^2+a(1)+b=5 \\ f(-1)=(-1)^2+a(-1)+b=5\]

Answer 3

\[a+b=4 \\ -a+b=4\]

Answer 4

is \(f(1)\) for \((x-1)\)

Answer 5

yep

Answer 6

f(-1) is for dividing by x-(-1)

Answer 7

what is this method called

Answer 8

remainder theorem

Answer 9

\[\frac{P(x)}{x-1}=Q(x)+\frac{R}{x-1} \\ P(x)=Q(x)(x-1)+R \\ \text{ plugin 1} \\ P(1)=Q(1)(1-1)+R \\ P(1)=Q(1)(0)+R \\ P(1)=0+R \\ P(1)=R\]

Answer 10

where our R=5 here

Answer 11

and we can then divide P(x) by (x+1)
and plug in -1
and you see we will get R there too

Answer 12

thnx

Answer 13

the previus one is easy