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# 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

#### 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

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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.
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