How do you use the remainder theorem to determine the remainder when the polynomial $\frac{2 t - 4 t^{3} - 3 t^{2}}{t - 2}$ $(2t-4t^3-3t^2)/(t-2)$ ?

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Answer 1 · 2017-03-31T06:30:57

The remainder is $- 40$ $-40$

According to remainder theorem, when a polynomial function $f (x)$ $f(x)$ is divided by $(x - a)$ $(x-a)$ , the remainder is $f (a)$ $f(a)$ .

Here we have the function $f (t) = - 4 t^{3} - 3 t^{2} + 2 t$ $f(t)=-4t^3-3t^2+2t$ , which is divided by $t - 2$ $t-2$ and hence the remainder is

$f (2) = - 4 \times 2^{3} - 3 \times 2^{2} + 2 \times 2$ $f(2)=-4xx2^3-3xx2^2+2xx2$

$= - 4 \times 8 - 3 \times 4 + 4$ $=-4xx8-3xx4+4$

$= - 32 - 12 + 4 = - 40$ $=-32-12+4=-40$

One can check it too

$f (t) = - 4 t^{3} - 3 t^{2} + 2 t$ $f(t)=-4t^3-3t^2+2t$

= $- 4 t^{2} (t - 2) - 8 t^{2} - 3 t^{2} + 2 t = - 4 t^{2} (t - 2) - 11 t^{2} + 2 t$ $-4t^2(t-2)-8t^2-3t^2+2t=-4t^2(t-2)-11t^2+2t$

(we have subtracted $8 t^{2}$ $8t^2$ to compensate for $- 4 t^{2} \times (- 2) = 8 t^{2}$ $-4t^2xx(-2)=8t^2$ )

= $- 4 t^{2} (t - 2) - 11 t (t - 2) - 22 t + 2 t = - 4 t^{2} (t - 2) - 11 t (t - 2) - 20 t$ $-4t^2(t-2)-11t(t-2)-22t+2t=-4t^2(t-2)-11t(t-2)-20t$

= $- 4 t^{2} (t - 2) - 11 t (t - 2) - 20 (t - 2) - 40$ $-4t^2(t-2)-11t(t-2)-20(t-2)-40$

= $(- 4 t^{2} - 11 t - 20) (t - 2) - 40$ $(-4t^2-11t-20)(t-2)-40$

i.e. quotient is $(- 4 t^{2} - 11 t - 20)$ $(-4t^2-11t-20)$ and remainder is $- 40$ $-40$

How do you use the remainder theorem to determine the remainder when the polynomial 2t−4t3−3t2t−2(2t-4t^3-3t^2)/(t-2)?