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Use the remainder theorem to show that (x+1) is a factor of f(x)=2x^3-7x^2-5x+4

Use the remainder theorem to show that (x+1) is a factor of f(x)=2x^3-7x^2-5x+4-example-1
User CatBusStop
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well, the remainder theorem says that if the polynomial f(x) has a factor of (x-a), then if we just plug in the "a" in f(x) it'll gives a remainder, assuming (x-a) is indeed a factor, then that remainder must be 0, so if f(a) = 0 then indeed (x-a) is a factor of f(x). After all that mumble jumble, let's proceed, we have (x+1), that means [ x - (-1) ], so if we plug in -1 in f(x), we should get 0, or f(-1) = 0, let's see if that's true.


\begin{array}{llrll} f(-1)&=&2(-1)^3-7(-1)^2-5(-1)+4\\\\ &&2(-1)-7(1)-5(-1)+4\\\\ &&-2-7+5+4\\\\ &&-9+9\\ &&0&~~\checkmark \end{array}

User Erikvold
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