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RD Chapter 6 Factorisation of Polynomials Ex VSAQS Solutions

Question - 1 : - Define zero or root of a polynomial.

Answer - 1 : - A real number a is a zero or root of a polynomial f(x) if f(α) = 0

Question - 2 : - If x = 12 is a zero of the polynomial f(x) =  8x3 +ax2 – 4x + 2, find the value of a.

Answer - 2 : -


Question - 3 : - Write the remainder whenthe polynomial f(x) = x3+x2-3x + 2is divided by x + 1.

Answer - 3 : -

f(x) = x3+x2-3x + 2
Let x + 1 = 0, then x = -1
∴Remainder =(-1)
Now,f(-1) = (-1)3 + (-1)2 – 3(-1) + 2
= -1 + 1+ 3 + 2 = 5
∴Remainder = 5

Question - 4 : - Find the remainder when x3 +4x2 + 4x – 3 is divided by x.

Answer - 4 : -

f(x) = x3 + 4x2 + 4x – 3
Dividing f(x) by x, we get
Let x = 0, then
f(x) = 0 + 0 + 0 – 3 = -3
∴  Remainder= -3

Question - 5 : - If x + 1 is a factor of x3 +a, then write the value of a.

Answer - 5 : -

Let f(x) = x3 + a
∴ x + 1is a factor of fx)
Let x + 1 = 0
⇒ x = -1
∴ f(-1)= x3 + a
= (-1)3 + a = -1 + a
∴ x + 1 is a factor
∴ Remainder = 0
∴ -1 + a = 0 ⇒ a = 1
Hence a = 1

Question - 6 : - If f(x) = x4-2x3 +3x2 – ax – b when divided by x – 1, the remainder is 6, thenfind the value of a + b.

Answer - 6 : -

f(x) = x4 – 2x3 + 3x2 –ax – b
Dividing f(x) by x – 1, the remainder = 6
Now let x – 1 = 0, then x = 1
∴ f(1) = (1)4 – 2(1)3 + 3(1)2 -ax1-b
= 1 -2 + 3-a-b = 2-a-b
∴Remainder = 6
∴ 2 – a– b = 6  ⇒ a + b= 2 – 6 = -4
Hence a + b = -4

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