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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’. - Mathematics

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प्रश्न

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’. 

योग

उत्तर

Let f(x) = 2x3 – 7x2 + ax – 6

x – 2 = 0 `\implies` 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) = x3 – 8x2 + (2a + 1)x – 16 

When g(x) is divided 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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अध्याय 8: Remainder and Factor Theorems - Exercise 8 (A) [पृष्ठ १०९]

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सेलिना Mathematics [English] Class 10 ICSE
अध्याय 8 Remainder and Factor Theorems
Exercise 8 (A) | Q 15 | पृष्ठ १०९

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