Question

By remainder theorem find the remainder when p(y)=4y^3-12y^2+5y-4 is divided by g(y)=2y-1

Answer 1

Step-by-step explanation:

Given

P(y) = 4y^3-12y^2+5y-4

g(y) = 2y-1

by remainder theorem

2y-1 = 0

2y = 1

y = 1/2

therefore,

p(1/2) = 4×(1/2)^3-12(1/2)^2+5(1/2)-4

p(1/2) = 4× 1/8 - 12×1/4 + 5/2 -4

p(1/2) = 1/2 - 3 +5/2 -4

p(1/2) = -8/2 = -4

Hence, (-4) is the remainder when p(y) = 4y^3-12y^2+5y-4.

Answer 2

Answer:

The reminder = -4

Step-by-step explanation:

Given p(y) = $4y^{3} -12y^{2} + 5y- 4$

          g(y) = 2y- 1    

and p(y) is divided by  g(y)

take 2y-1 =0 ⇒ 2y =1 ⇒  y = 1/2  

Here we need to find reminder by using Reminder theorem

Reminder Theorem

     In a polynomial f(x) is divided by  (x - a) then the reminder will be f(2)  

⇒ from data p(y) is divided by g(y) them the reminder is p(1/2)

⇒ p(1/2) =  $4(\frac{1}{2})^{3} - 12(\frac{1}{2} )^{2} + 5(\frac{1}{2} ) -4$

             =  $4(\frac{1}{2})( \frac{1}{2})(\frac{1}{2}) - 12(\frac{1}{2} ) (\frac{1}{2} ) + \frac{5}{2} -4$

             =  $\frac{1}{2} -3 + \frac{5}{2} -4$

             =  $\frac{1-6+5-8}{2}$ = $-\frac{8}{2}$ = - 4