Question

without actual division,prove that
$( {x}^{2} - 4 {x}^{2} + 12x - 9)$
is exactly divisible by
$(x + 3)$

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Answer 1

Step-by-step explanation:

$p(x) = {x}^{2} - 4 {x}^{2} - 12x - 9 = - 3 {x}^{2} + 12x - 9 \\ zero \: of \: (x + 3) \: is \: - 3. \\ p(x) = 0 \\ p(x) = - 3 {( - 3)}^{2} - 12 ( - 3 ) - 9 \\ = - 27 + 36 - 9 \\ = - 36 + 36 \\ p(x) = 0 \\ hence \: proved$

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