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RD Chapter 6 Factorisation of Polynomials Ex 6.3 Solutions

Question - 1 : - f(x) = x3 + 4x2 – 3x + 10, g(x) = x + 4

Answer - 1 : -


Question - 2 : - f(x) – 4x4 – 3x3 – 2x2 + x – 7, g(x) = x – 1

Answer - 2 : -


Question - 3 : - f(x) = 2x4 – 6X3 + 2x2 – x + 2, ,g(x) = x + 2

Answer - 3 : -


Question - 4 : - f(x) = 4x3 – 12x2 + 14x – 3, g(x) = 2x – 1

Answer - 4 : -


Question - 5 : - f(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – 2x

Answer - 5 : -


Question - 6 : - f(x) = x4 – 3x2 + 4, g(x) = x – 2

Answer - 6 : -


Question - 7 : -

Answer - 7 : -


Question - 8 : -

Answer - 8 : -


Question - 9 : - If the polynomials 2x3 + ax2 +3x – 5 and x3 + x2 – 4x + a leave the sameremainder when divided by x – 2, find the value of a.

Answer - 9 : -

Letf(x) = 2x3 + ax2 + 3x – 5
g(x) = x3+x2-4x + a
q(x) = x –  2 ⇒ x-2 =0  ⇒x = 2
∴Remainder =f(2) = 2(2)3 + a(2)2 + 3 x 2-5
= 2 x 8 4-a x 4 + 3 x 2-5
= 16 + 4a + 6 – 5
= 4a +17
and g(2) = (2)3 + (2)2 -4×2 + a
= 8 + 4 – 8 + a = a + 4
∵ In both cases, remainder are same
∴ 4a + 17 = a + 4
⇒ 4a – a = 4 – 17 ⇒ 3a = -13
⇒ a= −133
Hence a = −133

Question - 10 : - If the polynomials ax3 + 3x2 –13 and 2x3 – 5x + a, when divided by (x – 2), leave the sameremainders, find the value of a.

Answer - 10 : -

Letp(x) = ax3 + 3x2 – 13
q(x) = 2x3 –5x + a
and divisor g(x) = x – 2
x-2 = 0
⇒ x = 2
∴Remainder = p(2) = a(2)3 + 3(2)2 – 13
= 8a + 12 – 13 = 8a – 1
and q( 2) = 2(2)3 – 5×2 + a=16-10 + a
= 6 + a
∵ In each case remainder is same
∴ 8a – 1= 6 + a
8a – a = 6 + 1
⇒ 7a = 7
⇒ a= 77= 1
∴ a = 1

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