Question

using remainder theorem factorise the polynomial 2(x)4-7(x)3-13(x)2+63x-45
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Answer 1

$\purple{\bf{45 ⇒ ±1,±3,±5,±9,±15,±45}}$

$\purple{\bf{if we put x = 1 in p(x)}}$

p(1) =2(1)4 - 7(1)3 - 13(1)2 + 63(1) - 45

2 - 7 - 13 + 63 - 45 = 65 - 65 = 0

∴ x = 1 or x - 1 is a factor of p(x).

Similarly, if we put x = 3 in p(x)

p(3) = 2(3)4 - 7(3)3 - 13(3)2 + 63(3) - 45

162 - 189 - 117 + 189 - 45 = 162 - 162 = 0

Hence, x = 3 or x - 3 = 0 is the factor of p(x).

p(x) = 2x4 - 7x3 - 13x2 + 63x - 45

∴ p(x) = 2x3 (x - 1) -5x2 (x - 1) - 18(x - 1) + 45(x - 1)

2x4 - 2x3 (x - 1) - 5x2 - 18x2 + 18x + 45x - 54

⇒ p(x) = (x - 1)(2x3 - 5x2 - 18x + 45)

⇒ p(x) = (x - 1)(2x3 - 5x2 - 18x + 45)

⇒ p(x) = (x - 1)[2x2 (x - 3) + x(x - 3) - 15(x - 3)]

⇒ p(x) = (x - 1)[2x3 - 6x2 + x2 - 3x - 15x + 45]

⇒ p(x) = (x - 1)(x - 3)(2x2 + x - 15)

⇒ p(x) = (x - 1)(x - 3)(2x2 + 6x - 5x - 15)

⇒ p(x) = (x - 1)(x - 3)[2x(x + 3) - 5(x + 3)]

⇒ p(x) = (x - 1)(x - 3)(x + 3)(2x - 5)