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Polynomials

Short Answer Type

121. Factorise
2y³ + y² - 2y - 1

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122. Use Factor Theorem to verify that x + a is a factor of xⁿ + aⁿ for any odd positive integer n.

363 Views

123. Determine the value of ‘b’ for which the polynomial 5x³ - x² + 4x + b is divisible by 1 - 5x.

582 Views

124. Find the values of a and b so that (x + 1) and (x - 1) are factors of x⁴ + ax³ - 3x² + 2x + b.

183 Views

125. For what value of a the polynomial 2x³ + ax² + 11x + a + 3 is exactly divisible by 2x - 1?

262 Views

126. Without actual division prove that x⁴ + 2x³ - 2x² + 2x - 3 is exactly divisible by x² + 2x - 3.

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127. Determine whether (3x — 2) is a factor of 3x³ + x² - 20x + 12?

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128. Factorise: x³ - 6x² + 11x - 6.

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129. If both (x - 2) and $open parentheses straight x minus 1 half close parentheses$ are factors of px² + 5x + r, show that p = r.

Let f(x) = px² + 5 x + r
If (x - 2) is a factor of f (x), then by factor theorem
f(2) = 0 | x - 2 = 0 ⇒ x = 2
⇒ p(2)² + 5(2) + r = 0
⇒ 4p + r + 10 = 0 ...(1)

If $open parentheses straight x minus 1 half close parentheses$ is a factor of f (x), then by factor theorem,

$straight f open parentheses 1 half close parentheses equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space vertical line space straight x minus 1 half equals 0 rightwards double arrow straight x equals 1 half space rightwards double arrow space straight x equals 1 half rightwards double arrow space space space straight p open parentheses 1 half close parentheses squared plus 5 open parentheses 1 half close parentheses plus straight r equals 0 rightwards double arrow space space space space straight p over 4 plus 5 over 2 plus straight r equals 0 rightwards double arrow space space space straight p plus 4 straight r plus 10 equals 0 space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis$