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Polynomials

Short Answer Type

111. Find the value of k, if x - 1 is a factor of p(x) in each of the following cases:
p(x) = 2x²+ kx +

square root of 2

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112. Find the value of k, if x - 1 is a factor of p(x) in each of the following cases:

p(x) = kx² -

square root of 2 to the power of straight x end root plus 1

160 Views

113. Find the value of k, if x - 1 is a factor of p(x) in each of the following cases:

p(x)=kx² - 3x + k

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114. Factorise
12x² - 7x + 1

150 Views

115. Factorise
2x² + 7x + 3

130 Views

116. Factorise
6x² + 5x - 6

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117. Factorise
3x² - x - 4

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118. Factorise
x³ - 2x² - x + 2

x³ - 2x² - x + 2
Let p(x) = x³ - x² - x + 2
By trial, we find that
p(1) = (1)³ - 2(1)² - (1) + 2
= 1 - 2 - 1 + 2 = 0
∴ By Factor Theorem, (x - 1) is a factor of p(x).
Now,
x³ - 2x² - x + 2 = x²(x - 1) - x(x - 1) - 2(x - 1)
= (x - 1)(x² - x - 2)
= (x - 1)(x² - 2x + x - 2)
= (x - 1){x(x - 2) + 1(x - 2)}
= (x - 1)(x - 2)(x + 1).