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प्रश्न
(3x + 5) is a factor of the polynomial (a – 1)x3 + (a + 1)x2 – (2a + 1)x – 15. Find the value of ‘a’, factorise the given polynomial completely.
उत्तर
Let f(x) = (a – 1)x3 + (a + 1)x2 – (2a + 1)x – 15
It is given that (3x + 5) is a factor of f(x).
∴ Remainder = 0
`f((-5)/3) = 0`
`(a - 1)(-5/3)^3 + (a + 1)((-5)/3)^2 - (2a + 1)((-5)/3) - 15 = 0`
`(a - 1)((-125)/27) + (a + 1)(a + 1)(25/9) - (2a + 1)((-5)/3) - 15 = 0`
`(-125(a-1) + 75(a+1) + 45(2a+1) - 405)/27 = 0`
–125a + 125 + 75a + 75 + 90a + 45 – 405 = 0
40a – 160 = 0
40a = 160
a = 4
∴ f(x) = (a – 1)x3 + (a + 1)x2 – (2a + 1)x – 15
= 3x3 + 5x2 – 9x – 15
x2 – 3
`3x + 5")"overline(3x^3 + 5x^2 - 9x - 15)`
3x3 + 5x2
– 9x – 15
– 9x – 15
0
∴ 3x3 + 5x2 – 9x – 15 = (3x + 5)(x2 – 3)
= `(3x + 5)(x + sqrt(3))(x - sqrt(3))`
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