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anonymous

  • one year ago

Factor Theorem and Remainder Theorem Help?

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  1. anonymous
    • one year ago
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    I can't wrap my head around these subjects!

  2. freckles
    • one year ago
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    \[f(x)=x^2+x-1 \\ \text{ say we want to find } f(2)\] 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. |dw:1438650916009:dw| 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.

  3. freckles
    • one year ago
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    now you know (x-3) is a factor of x^2-9 |dw:1438651586427:dw| the remainder is 0 So if (x-c) 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 x-c is a factor of f(x)

  4. freckles
    • one year ago
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    another 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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