Question

guyssss please answerrrr!


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Answer 1

Answer:

$f(t) = {t}^{2} -4t+3 \\ and \: given \: that \: the \: zeroes \:are \: \alpha \: \\ and \: \beta \\ {t}^{2} - 4t + 3 \\ = {t}^{2} - 3t - t + 3 \\ = t(t - 3) - 1(t - 3) \\ = (t - 1)(t - 3) \\ t = 3 \: or \: t = 1 \\ \alpha = 1 \: and \: \beta = 3$

$now \\ { \alpha }^{4} { \beta }^{3} + { \alpha }^{3} { \beta }^{4} \\ = {(1)}^{4} {(3)}^{3} + { (1)}^{3} {(3)}^{4 } \\ = 27 + 81 \\ = 108$