Question

by actual division find the quotient and the remainder when
(i) f(y) =3y³+6y²+7y-2 id divided by 3y -1

(ii)q(t) =2t⁴+3t³-2t²+3t+6 is divided by ( t + 2)

NO SPAMMING
solve with solution......​

Answer 1

➡️lets solve for f

$fy = \frac{ {3y}^{3} + {6y}^{2} + 7y - 2 }{3y - 1}$

Multiply both sides by 3y-1

$3f {y}^{2} - fy = {3y}^{3} + 6 {y}^{2} + 7y - 2$

Factor out variable f

$f( {3y}^{2} - y) = {3y}^{3} + {6y}^{2} + 7y - 2$

Divide both sides by 3y^2-y

$\frac{f( {3y}^{2} - y) }{ {3y}^{2} - y} = \frac{ {3y}^{2} + {6y}^{2} + 7y - 2}{ {3y}^{2} - y}$

$f = \frac{ {3y}^{3} + {6y}^{2} + 7y - 2 }{3 {y}^{2} - y}$

➡️Lets solve for q

$qt = \frac{ {2t}^{4} + {3t}^{3} - {2t}^{2} + 3t + 6 }{t + 2}$

Multiply both sides by t+2

${qt}^{2} + 2qt = {2t}^{4} + {3t}^{3} - {2t}^{2} + 3t + 6$

factor out variable q.

$q( {t}^{2} + 2t) = {2t}^{4} + {3t}^{3} - {2t}^{2} + 3t + 6$

Divide both sides by t^2+2t

$\frac{q( {t}^{2} + 2t) }{ {t}^{2} 2t} = \frac{ {2t}^{4 + 3 {t}^{3} } - 2 {t}^{2} + 3t + 6 }{ {t}^{2} + 2t }$

$q = \frac{ {2t}^{4} + {3t}^{3} + {2t}^{2} + 3t + 6}{ {t}^{2} + 2t }$