What is the remainder when 3t^32t+2 is divided by 3t+1?

answers:
Ian H say: Dividing a polynomial f(x) by x – a, (with degree 1), the result takes the form
f(x) = (x – a)g(x) + c, (where remainder c has degree 0, i.e. it is a constant).
Now if we let x = a, we get a consequence known as the Remainder Theorem
f(a) = c
We find the remainder of dividing f(x) by x – a, by substituting x = a into f(x)
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What is the remainder when 3t^3  2t + 2 is divided by 3t + 1?
When 3t + 1 = 0, t = 1/3, so that is what we substitute;
f(1/3) = 3(1/3)^3  2(1/3) + 2 = = 1/9 + 2/3 + 2 = 23/9
Extra note: f(x) = (3t^3  2t  5/9) + 23/9
and dividing by 3t + 1 gives t^2 – t/3 with remainder 23/9

sepia say: (3t^3  2t + 2) / (3t + 1)
Quotient and remainder:
3 t^3  2 t + 2 = (t^2  t/3  5/9) × (3 t + 1) + 23/9

khalil say: 3t+1 = 0
t = 1/3
r = f(1/3) = 3(1/3)³ 2(1/3) +2
r = 23/9

la console say: (3t³  2t + 2) / (3t + 1)
First term: 3t³/3t = t²
t².(3t + 1) = 3t³ + t²
Rest:
= (3t³  2t + 2)  (3t³ + t²)
= 3t³  2t + 2  3t³  t²
=  t²  2t + 2
Second term:  t²/3t =  (1/3).t
 (1/3).t.(3t + 1) =  t²  (1/3).t
Rest:
= ( t²  2t + 2)  [ t²  (1/3).t]
=  t²  2t + 2 + t² + (1/3).t
=  (5/3).t + 2
Third term:  (5/3).t/3t =  5/9
 (5/9).(3t + 1) =  (5/3).t  (5/9)
Rest:
= [ (5/3).t + 2]  [ (5/3).t  (5/9)]
=  (5/3).t + 2 + (5/3).t + (5/9)
= 2 + (5/9)
= 23/9 ← this is the remainder
3t³  2t + 2 = [(3t + 1).[t²  (1/3).t  (5/9)] + (23/9)
(3t³  2t + 2) / (3t + 1) = t²  (1/3).t  (5/9) + { 23/[9.(3t + 1)] }
(3t³  2t + 2) / (3t + 1) = t²  (1/3).t  (5/9) + [23/(27t + 9)]

nbsale say: You get a quotient of t^2 t/3 5/9 and a remainder of 23/9

llaffer say: If dividing by (3t + 1), the remainder is the same as if t = 1/3 (what makes that expression a zero).
So solve for f(1/3) and that's your remainder:
f(t) = 3t³  2t + 2
f(1/3) = 3(1/3)³  2(1/3) + 2
f(1/3) = 3(1/27)  2(1/3) + 2
f(1/3) = 1/9 + 2/3 + 2
f(1/3) = 1/9 + 6/9 + 18/9
f(1/3) = 23/9
I get the same as Morningfox.

tiyt say: try an online cal

Morningfox say: I get a remainder of 23/9.
