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 

${\displaystyle f\left(\frac{-5}{3}\right)=0}$ 

${\displaystyle (a-1){(-\frac{5}{3})}^3+(a+1){\left(\frac{-5}{3}\right)}^2-(2a+1)\left(\frac{-5}{3}\right)-15=0}$

${\displaystyle (a-1)\left(\frac{-125}{27}\right)+(a+1)(a+1)\left(\frac{25}{9}\right)-(2a+1)\left(\frac{-5}{3}\right)-15=0}$ 

${\displaystyle \frac{-125(a-1)+75(a+1)+45(2a+1)-405}{27}=0}$ 

–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
${\displaystyle 3x+5\text{)}\overline{3x^3+5x^2-9x-15}}$
             3x³ + 5x²                    
                               – 9x – 15
                               – 9x – 15   
                                      0

∴ 3x³ + 5x² – 9x – 15 = (3x + 5)(x² – 3) 

= ${\displaystyle (3x+5)(x+\sqrt{3})(x-\sqrt{3})}$