Question

(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.

Answer 1

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))`