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anonymous
 one year ago
Factor Theorem and Remainder Theorem Help?
anonymous
 one year ago
Factor Theorem and Remainder Theorem Help?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can't wrap my head around these subjects!

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f(x)=x^2+x1 \\ \text{ say we want to find } f(2)\] Well we could that by inputting 2 \[f(2)=2^2+21 \\ f(2)=4+21 \\ f(2)=61 \\ f(2)=5 \] Guess what we will also see 5 is the remainder of f(x) divided by (x2) Why? Well let's see why. dw:1438650916009:dw so this means we have \[\frac{f(x)}{x2}=x+3+\frac{5}{x2} \\ \text{ multiply both sides by } (x2) \text{ gives:} \\ f(x)=(x+3)(x2)+5 \text{ as you see pluggin in 2 for } x \text{ gives } \\ f(2)=(2+3)(22)+5 \\ f(2)=0+5=5 \\ f(2)=5 \] This is the remainder theorem Let f(x) be a polynomial. When f(x) is divided by xc the remainder is f(c). That is... \[\frac{f(x)}{xc}=Q(x)+\frac{R(x)}{xc} \\ Q(x) \text{ is the quotient as you seen in the example } \\ R(x) \text{ will be a constant since we are dividing by a linear function } (xc) \\ \text{ so let me rewrite this a bit } \\ \frac{f(x)}{xc}=Q(x)+\frac{R}{xc} \\ \text{ so multiplying both sides by } (xc) \\ f(x)=Q(x) \cdot (xc)+R \\ \text{ as you see inputting } c \text{ for } x \text{ gives } \\ f(c)=Q(c) \cdot (cc)+R \\ f(c)=0+R \\ f(c)=R \\ \text{ see } f(c) \text{ is the remainder } R \text{ when } \\ f(x) \text{ is divide by } (xc)\] Note: Q(x) is a polynomial too.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2now you know (x3) is a factor of x^29 dw:1438651586427:dw the remainder is 0 So if (xc) is a factor of f(x) then the remainder (or f(c) or R as we called it above) is 0 or the other way around if f(c)=0 then xc is a factor of f(x)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2another example: \[\frac{x^2+5x+6}{x+2}=x+3+\frac{0}{x+2} \text{ note: notice: remainder is 0} \\\ \text{ now this is the same as } \\ \frac{x^2+5x+6}{x+2}=x+3 \\ \text{ multiply both sides by } (x+2) \\ x^2+5x+6=(x+3)(x+2) \\ \text{ as we see } (x+2) \text{ is totally a factor of } x^2+5x+6 \\ \text{ and we also see the remainder was 0 when } x^2+5x+6 \text{ was } \\ \text{ divided by } x+2 \\ \text{ also notice } f(x)=x^2+5x+6 \text{ gives } f(2)=0\]
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