Question 82 (iv)

Subtract

$3 t^{4} - 4 t^{3} + 2 t^{2} - 6 t + 6 f r o m - 4 t^{4} + 8 t^{3} - 4 t^{2} - 2 t + 11$

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Solution

The required difference is given by
$(- 4 t^{4} + 8 t^{3} - 4 t^{2} - 2 t + 11) - (3 t^{4} - 4 t^{3} + 2 t^{2} - 6 t + 6)$
$- 4 t^{4} + 8 t^{3} - 4 t^{2} - 2 t + 11 - 3 t^{4} + 4 t^{3} - 2 t^{2} + 6 t - 6$
$= (- 4 t^{4} - 3 t^{4}) + (8 t^{3} + 4 t^{3}) + (- 4 t^{2} - 2 t^{2}) + (- 2 t + 6 t) + (11 - 6)$
[grouping like terms]
$= - 7 t^{4} + 12 t^{3} - 6 t^{2} + 4 t + 5$

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Subtract

$3 t^{4} - 4 t^{3} + 2 t^{2} - 6 t + 6 f r o m - 4 t^{4} + 8 t^{3} - 4 t^{2} - 2 t + 11$

Is 2t+1 a factor of 4 $t^{3}$ +4 $t^{2}$ -t-1?

g (t) = 4 t^{3} + 4 t^{2} - t - 1

f (t) = 2 t + 1

q

R

q (t) = 4 t^{3} + 4 t^{2} - t - 1

(2 t + 1)

Find the remainder when q(t) = 4t³ + 4t² – t - 1 is divided by 2t + 1

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