Question

Factor Theorem and Remainder Theorem Help?

Answer 1

I can't wrap my head around these subjects!

Answer 2

\[f(x)=x^2+x-1 \\ \text{ say we want to find } f(2)\]
Well we could that by inputting 2
\[f(2)=2^2+2-1 \\ f(2)=4+2-1 \\ f(2)=6-1 \\ f(2)=5 \]
Guess what we will also see 5 is the remainder of f(x) divided by (x-2)
Why?
Well let's see why.

so this means we have
\[\frac{f(x)}{x-2}=x+3+\frac{5}{x-2} \\ \text{ multiply both sides by } (x-2) \text{ gives:} \\ f(x)=(x+3)(x-2)+5 \text{ as you see pluggin in 2 for } x \text{ gives } \\ f(2)=(2+3)(2-2)+5 \\ f(2)=0+5=5 \\ f(2)=5 \]
This is the remainder theorem
Let f(x) be a polynomial. When f(x) is divided by x-c the remainder is f(c).
That is...
\[\frac{f(x)}{x-c}=Q(x)+\frac{R(x)}{x-c} \\ Q(x) \text{ is the quotient as you seen in the example } \\ R(x) \text{ will be a constant since we are dividing by a linear function } (x-c) \\ \text{ so let me rewrite this a bit } \\ \frac{f(x)}{x-c}=Q(x)+\frac{R}{x-c} \\ \text{ so multiplying both sides by } (x-c) \\ f(x)=Q(x) \cdot (x-c)+R \\ \text{ as you see inputting } c \text{ for } x \text{ gives } \\ f(c)=Q(c) \cdot (c-c)+R \\ f(c)=0+R \\ f(c)=R \\ \text{ see } f(c) \text{ is the remainder } R \text{ when } \\ f(x) \text{ is divide by } (x-c)\]
Note: Q(x) is a polynomial too.