ICSE Class 10CISCE
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

The Polynomials 2x3 – 7x2 + Ax – 6 and X3 – 8x2 + (2a + 1)X – 16 Leaves the Same Remainder When Divided by X – 2. Find the Value of ‘A’. - ICSE Class 10 - Mathematics

Login
Create free account


      Forgot password?

Question

The polynomials 2x3 – 7x2 + ax – 6 and x3 – 8x2 + (2a + 1)x – 16 leaves the same remainder when divided by x – 2. Find the value of ‘a’. 

Solution

Lef f(x)=`2x^3-7x^2+ax-6`  

x-2=0 ⇒  x=2 

When f (x) is divided by (x-2), remainder = f(2) 

∴ `f(2)=2(2)^3-7(2)^2+a(2)-6` 

          =`16-28-2a-6`    

          = 2a-18 

Let `g(x)=x^3-8x^2+(2a+1)x-16` 

When g (x) is dividend by (x-2), remainder=g(2) 

∴ `g(2)=(2)^3-8(2)^2+(2a+1)(2)-16` 

          = 8-32+4a+2-16 

        = 4a-38  

By the given condition, we have: 

f(2)=g(2) 

2a-18=4a-38 

4a-2a=38-18  

2a=20 

a=10 

Thus, the value of a is 10

  

  Is there an error in this question or solution?

Video TutorialsVIEW ALL [2]

Solution The Polynomials 2x3 – 7x2 + Ax – 6 and X3 – 8x2 + (2a + 1)X – 16 Leaves the Same Remainder When Divided by X – 2. Find the Value of ‘A’. Concept: Remainder Theorem.
S
View in app×