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Mathematics

(3x + 5) is a factor of the polynomial (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15. Find the value of 'a'. For this value of 'a', factorise the given polynomial completely.

Factorisation

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Answer

3x + 5 = 0 ⇒ x = -53\dfrac{5}{3}

Since, (3x + 5) is a factor of the polynomial (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15. Substituting x = -53\dfrac{5}{3} in (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15, remainder = 0.

⇒(a−1)(−53)3+(a+1)(−53)2−(2a+1)(−53)−15=0⇒(a−1)(−12527)+(a+1)(259)+10a+53−15=0⇒−125a+12527+25a+259+10a+53=15⇒−125a+125+75a+75+90a+4527=15⇒40a+245=405⇒40a=160⇒a=4.\Rightarrow (a - 1)\Big(-\dfrac{5}{3}\Big)^3 + (a + 1)\Big(-\dfrac{5}{3}\Big)^2 - (2a + 1)\Big(-\dfrac{5}{3}\Big) - 15 = 0 \\[1em] \Rightarrow (a - 1)\Big(-\dfrac{125}{27}\Big) + (a + 1)\Big(\dfrac{25}{9}\Big) + \dfrac{10a + 5}{3} - 15 = 0 \\[1em] \Rightarrow \dfrac{-125a + 125}{27} + \dfrac{25a + 25}{9} + \dfrac{10a + 5}{3} = 15 \\[1em] \Rightarrow \dfrac{-125a + 125 + 75a + 75 + 90a + 45}{27} = 15 \\[1em] \Rightarrow 40a + 245 = 405 \\[1em] \Rightarrow 40a = 160 \\[1em] \Rightarrow a = 4.

Substituting a = 4 in (a - 1)x3 + (a + 1)x2 - (2a + 1)x - 15,

⇒ (4 - 1)x3 + (4 + 1)x2 - (2(4) + 1)x - 15

⇒ 3x3 + 5x2 - 9x - 15

⇒ x2(3x + 5) - 3(3x + 5)

⇒ (x2 - 3)(3x + 5)

⇒ (x−3)(x+3)(3x+5)(x - \sqrt{3})(x + \sqrt{3})(3x + 5)

Hence, a = 4 and 3x3 + 5x2 - 9x - 15 = (x−3)(x+3)(3x+5)(x - \sqrt{3})(x + \sqrt{3})(3x + 5).

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